SECOND 
COURSE   IN   ALGEBRA 


BY 
WEBSTER   WELLS,    S.B. 

AUTHOR    OF    A    SERIES    OP    TEXTS    ON    MATHEMATICS 
AND 

WALTER    W.    HART,    A.B. 

ASSISTANT    PROFESSOR    OK    MATHEMATICS,    UNIVERSITY    OF   WISCONSIN 
COURSE    FOR    TH3    TRAINING    OF    TEACHERS 


D.    C.    HEATH    &    CO.,   PUBLISHERS 
BOSTON  NEW  YORK  CHICAGO 


iv  PREFACE 

* 
students  a  rather  complete  treatment  of  these  important  top- 
ics, even  though  they  may  not  care  to  cover  all  of  the  examples 
and  problems  of  the  text.  Special  attention  is  directed  to  the 
emphasis  placed  upon  getting  roots  of  quadratic  equations  in 
decimal  form.     (See  §  78.) 

(c)  Chapter  IX  is  an  elaboration  of  Chapter  XVI  of  the 
First  Year  Algebra.  The  advantage  of  postponing  the  topics 
in  this  chapter  until  well  into  the  third  semester  of  algebra  is 
apparent.  Isolating  these  topics  as  in  this  text  makes  it  pos- 
sible for  schools  that  desire  a  brief  course  to  omit  the  chapter 
altogether.  In  this  chapter,  topics  such  as  those  in  §§89  to 
95  find  a  logical  place. 

(d)  The  remaining  chapters  contain  the  topics  which  appear 
among  the  various  college  entrance  requirements  in  algebra. 
In  the  main,  all  colleges  enumerate  the  topics  considered  in 
Chapters  I  to  XIV.  The  other  topics  are  enumerated  by 
one  or  more  institutions.  Obviously,  teachers  will  need  to 
select  the  chapters  that  meet  the  needs  of  their  classes. 

(e)  Special  attention  is  directed  to  Chapters  XIII  and  XIV, 
on  Exponents  and  Radicals.  These  chapters,  while  of  con- 
siderable mathematical  interest,  probably  are  retained  in  sec- 
ondary courses  largely  because  they  appear  among  college 
entrance  requirements.  Usually  they  are  taught  in  the  first 
course.  They  are  so  difficult  that  students  acquire  there  little 
knowledge  and  very  little  skill  in  dealing  with  them.  Where 
preparation  for  an  examination  is  a  special  aim,  the  authors 
are  confident  that  instruction  on  these  topics  toward  the  latter 
part  of  the  third  semester  of  algebra  will  be  found  not  only 
more  pedagogical  but  more  timely.  The  examples  selected  for 
this  text  are  as  simple  as  possible.  More  difficult  examples 
on  Exponents  occur  in  the  supplementary  set  F.  Attention  is 
directed  to  the  practical  turn  given  to  radicals  in  §  126. 

(/)  Chapter  XXIV  contains  a  number  of  topics  that  will 
have  interest  for  some  teachers 


CONTENTS 


CHAPTER  PAGE 

I.  The  Fundamental  Operations          .         .         .               1 

II.  Special  Products  and  Factoring     •  .        .        .        14 

III.  Fractions .        .24 

IV.  Simple  Equations 33 

V.  Graphical  Representation        .        .        .        .        .    46 

VI.  Simultaneous  Linear  Equations           .        .        .        50 

VII.  Square  Root  and  Quadratic  Surds         .        .        .63 

VIII.     Quadratic  Equations 74 

IX.  Special  Products  and  Factoring     ....  100 

X.  Quadratic  Equations  having  Two  Variables  .       117 

XI.     Simultaneous  Equations 124 

XII.  The  Theory  of  Quadratic  Equations         .        .       135 

XIII.     Exponents 140 

XIV".     Radicals 150 

XV.     Logarithms    .        . 170 

XVI.     Progressions 187 

XVII.     The  Binominal  Theorem 202 

XVIII.  Ratio,  Proportion,  and  Variation       .        .        .      206 

XIX.  Graphical  Representation  of  Complex  Numbers  220 

XX.  Equations  in  the  Quadratic  Form      .         .         .       224 

XXI.  The  Binomial  Theorem  (Concluded)         .         .         .227 

XXII.  Permutations  and  Combinations  ....      231 

XXIII.  Determinants        ........  289 

XXTV.  Supplementary  Topics    ......      256 

INDEX 283 

v 


ALGEBRA 


I.   THE   FUNDAMENTAL   OPERATIONS 

UPON  ARITHMETICAL  NUMBERS 

1.  The  Number  System  of  Arithmetic  consists  of  the  integers 
and  the  common  fractions.  The  following  facts  about  arith- 
metical numbers  are  collected  here  for  reference : 

(a)  The  sum,  the  product,  and  the  quotient  of  two  arithmeti- 
cal numbers  is  always  an  arithmetical  number ;  the  difference 
between  two  such  numbers,  however,  is  an  arithmetical  num- 
ber only  when  the  minuend  is  greater  than  the  subtrahend. 
Division  by  zero  is  not  allowed. 

(b)  The  Associative  Law  of  Addition.  The  sum  of  three  or 
more  numbers,  addends,  is  the  same  in  whatever  manner  the 
addends  are  grouped.     Thus, 

a  +  b  +  c  =(a  +  b)  +  c  =  a  +  (b  +  c). 

(c)  The  Commutative  Law  of  Addition.  The  sum  of  two  or 
more  addends  is  the  same  in  whatever  manner  the  addends 
are  arranged.     Thus, 

a  +  b  +  c  =  a-\-c-\-b  =  c+b-\-a. 

(d)  The  Associative  Law  of  Multiplication.  The  product  of 
three  or  more  numbers,  factors,  is  the  same  in  whatever  man- 
ner the  factors  are  grouped.     Thus, 

a  •  b  •  c  =  (a  •  b)  •  c  =  a  •  (6  •  c). 
1 


2  ALGEBRA 

(e)  The  Commutative  Law  of  Multiplication.  The  product  of 
two  or  more  factors  is  the  same  in  whatever  manner  the  factors 
are  arranged.     Thus, 

a,'b-c  =  b-a-c  =  C'b'a. 

(/)  The  Distributive  Law  of  Multiplication.  If  the  sum  or 
the  difference  of  two  (or  more)  numbers  is  multiplied  by  a 
third  number,  the  product  may  be  found  by  multiplying  each 
of  the  numbers  separately  by  the  multiplier  and  connecting 
the  results  by  the  proper  signs.     Thus, 

a  (6  +  c  —  d)=  ab  -f  ac  —  ad. 

2.  Positive  and  Negative  Numbers.  To  avoid  the  difficulty 
that  subtraction  is  sometimes  impossible  in  the  system  of 
arithmetical  numbers,  the  negative  integers  and  fractions  are 
added  to  the  number  system.  The  combined  system  of  posi- 
tive and  negative  integers  and  fractions  is  the  System  of^ 
Rational  Numbers. 

The  following  facts  are  learned  in  the  first  course  in  algebra : 

(a)  To  every  arithmetical  number,  there  corresponds  a  posi- 
tive and  a  negative  number.  The  arithmetical  number  is  the 
Absolute  Value  of  each  of  the  corresponding  signed  numbers. 
Thus,  3  is  the  absolute  value  of  +  3  and  of  —  3. 

(b)  The  rational  scale  is : 

-5  -4  -3  -2  -1  0  +1  +2  -1-3         +4  45 

1 t— 1 1 1 1 1 H- 1 1 ■ 

The  fractions  correspond  to  points  between  the  points 
marked  by  the  integers.  Any  negative  number  precedes  all 
of  the  positive  numbers,  and  may,  therefore,  be  regarded  as 
being  less  than  any  positive  number ;  of  two  negative  numbers, 
the  one  having  the  greater  absolute  value  is  the  less. 

(c)  The  sum  of  any  positive  number  and  the  corresponding 
negative  number  is  zero.     Thus, 

(+3)  +  (-3)=0. 


THE   FUNDAMENTAL   OPERATIONS  3 

(d)  To  add  two  signed  numbers  having  the  same  sign,  add 
their  absolute  values  and  prefix  their  common  sign.     Thus, 

(_10)  +  (-2)=-12. 

(e)  To  add  two  numbers  having  unlike  signs,  find  the  differ- 
ence between  their  absolute  values  and  prefix  the  sign  of  the 
one  having  the  greater  absolute  value.     Thus, 

(+2)  +  (-9)=-7.         . 

(/)  To  subtract  one  signed  number  from  another,  change  the 

sign  of  the  subtrahend  and  add  the  result  to  the  minuend. 

Thus, 

(_3)-(-8)  =  (-3)  +  (+8)=  +  5. 

(g)  To  multiply  one  signed  number  by  another,  find  the 
product  of  their  absolute  values,  and  make  it  positive  if  the 
numbers  have  the  same  sign,  but  negative  if  they  have  unlike 
signs.     Thus, 

(-  3)  •  (+  9)=  -  27,  and  (-  5)  .  (-  16)=  +  80. 

(h)  To  divide  one  signed  number  by  another,  find  the  quo- 
tient of  their  absolute  values,  and  make  it  positive  if  the 
numbers  have  the  same  sign,  but  negative  if  they  have  unlike 
signs.     Thus, 

(+39)-*-(-3)=-13,  and  (-  45)-s-(+5)  =  -  9. 

3.  The  Fundamental  Operations  are '  addition,  subtraction, 
multiplication,  division,  involution,  and  evolution. 

Involution  is  the  process  of  finding  the  product  when  a  given 
number  is  used  as  a  factor  two  or  more  times.  The  number 
itself  is  called  the  Base;  the  result  is  called  a  Power  of  the 
•base ;  the  Exponent,  written  at  the  right  of  and  above  the  base, 
indicates  the  number  of  times  the  base  is  used  as  a  factor. 

Thus,  3*  =  3  •  3  •  3  •  3  =  81,  and  (2  6)8  =  (2  6)  •  (2  6)  •  (2  6)  =  8  63. 


4  ALGEBRA 

Evolution  is  the  process  of  finding  what  number  must'  be 
used  as  a  factor  a  specified  number  of  times  to  produce  a  given 
number  as  product.  The  given  number  is  called  the  Radicand ; 
the  result  is  called  a  Root  of  the  radicand ;  the  Radical  Sign, 
V  ,  with  the  proper  Index  denotes  the  desired  root. 

Thus,  VS  indicates  the  cube  root  of  8  ;  8  is  the  radicand,  3  is  the 
index  of  the  root,  and  the  root  itself  is  2,  for  23  =  8. 

Evolution  is  not  always  possible  in  the  system  of  rational 
numbers ;  thus,  there  is  no  rational  number  which  exactly  ex- 
presses the  square  root  of  2. 

Order  of  Operations.  In  a  sequence  of  the  fundamental  opera- 
tions on  numbers,  it  is  agreed  that  operations  under  radical 
signs  or  within  symbols  of  grouping  shall  be  performed  before 
all  others ;  that,  otherwise,  all  multiplications  and  divisions 
shall  be  performed  first,  proceeding  from  left  to  right,  and 
afterwards  all  additions  and  subtractions,  proceeding  again 
from  left  to  right. 

EXERCISE   1 

1.  Illustrate  by  arithmetical  examples  the  five  fundamental 
laws  in  paragraph  1. 

2.  Give  an  arithmetical  example  in  which  subtraction  is 
impossible.    Give  the  result  when  dealing  with  signed  numbers. 

3.  Subtraction  is  said  to  be  the  inverse  of  addition,  and 
division  of  multiplication.    What  is  meant  by  these  statements  ? 

4.  What  are  the  two  signed  numbers  corresponding  to 
J  ?  Give  their  sum,  their  difference,  their  product,  and  their 
quotient. 

5.  Perform  the  following  indicated  operations  : 
(«)    (+6)  +  (+10).  (e)    (-4)  +  (-3). 
(6)    (_4)  +  (+9).  (/)  (+6)  +  (-3). 

(c)  (_8)  +  (+6).  (fir)    (+6)-(-5). 

(d)  (  +  2)  +  (-3).  (A)   (_4)-(  +  2). 


THE   FUNDAMENTAL   OPERATIONS 

(9     ("5)-(-3).  ($    (-5).  (-9). 

0)     (-4)-(-8).  (o)     (+12).  (-3). 


(*)    (+3)-(+9).  (p)    (+25). 


(-5). 


(0     (+6).  (-5).  (?)    (_28)  +  (+4). 

(m)  (-7). (+8).  (r)     (-22)  + (-11). 

6.  What  is  the  value  of  34?  of  25?  of  63  ?  of  (-2)2?  of 
(_3)3?  of  (-4)2?  of  (-o)3? 

7.  What  is  the  sign  of  the  product  of  an  even  number  of 
negative  factors  ?  of  an  odd  number  ?  Give  an  example  of 
each. 

8.  What  is  the  sign  of  an  even  power  of  a  negative  num- 
ber ?  of  an  odd  power  ?     Give  an  example  of  each. 

9.  Read  the  following  product:  btfytfw.  Give  the  expo- 
nent of  each  of  the  literal  numbers.  What  is  the  value  of  the 
product  when  x  =  2,  y  =  —  3,  z  =  —  2,  and  w  =  5. 

10.  Find  the  values  of  the  following  for  the  values  of  x,  y, 
z,  and  w  given  in  Example  9 : 

(a)  Sx2.  (c)   ^  +  2/3.  (e)    x2-3x  +  4. 

(6)   x2y.  (d)  w2-z2.  (f)  z2-4z-4:. 

(g)  z2-2zw  +  w2.  (j)   a^+y4. 

(h)  **""**•  (k)  !°  +  l_I 

(t)    xA-2x*  +  x2-x  +  l.  hJ   z2     x2     yz 

Write  the  following  in  symbols  and  find  their  values : 
(I)     the  sum  of  the  squares  of  x  and  y. 
(m)  the  difference  between  the  cubes  of  w  and  x. 
(n)    twice  the  product  of  y  and  z,  diminished  by  three  times 
the  quotient  of  x  and  iv. 

11.  The  following  formulae  occur  in  applications  of  algebra. 
Those  marked  with  an  asterisk,  *,  occur  in  geometry ;  try  to 
recall  the  theorems  they  express.    Express  the  others  in  words. 


6  ALGEBRA 

(a)  *  A  =  |  a&.     Find  J.  when  a  =  25  and  6  =  40. 

(&)*  .4  =  £ft(&  +  6').     Find  .4  when/*  =  10,  6  =  30,  and  &'=20. 

(c)      ^4  =  pfl  +  ^-V     Find  A  when  P  =  5000,  r  =  5,  and 
t  =  4  *  ' 

(if)      F=  4  tt#3.     Find  F  when  tt  =  3±-  and  R  =  3. 

(e)  £  =  7t a/- •     Find  *  when  I  =  8  and  g  =  32. 

(/)     s  =  a£-fi^2.     Find  s  when  a  =50,  t  =  3,  and  #  =  32. 

(0)  *  7i2  =  a2  +  &2.     Find  h  when  a  =  6  and  b  =  8. 

(A)      S  =  ttI(R  +r).     Find  #  when  I  =  10,  72  =  5,  and  r  =  3. 

(f)  p  =  a2  -fft2  -  c2      Findi>  when  a  =  8?  b  =  9j  and  c  =  10 

( j)     a  =  ^+-  ch.     Find  .4  when  h  =  5  and  c  =  30. 

THE   FUNDAMENTAL   OPERATIONS 
UPON  NUMBER  EXPRESSIONS 
4.    (a)  An   Expression  is  a  symbol   for  a  number,  consist- 
ing  of    numerals    and    literal    numbers    connected    by    some 
or   all   of  the   signs  denoting   mathematical   operations;    as, 

^2  ajty  +  4  ^  -  iv\ 

(b)  A  Monomial  or  Term  is  an  expression  whose  parts  are  not 
connected  by  the  signs  +  or  —  ;  as,  6  rV£. 

(c)  A  Binomial  is  an  expression  having  two  terms. 

(d)  A  Trinomial -is  an  expression  having  three  terms. 

(e)  A  Polynomial  is  an  expression  having  more  than  one  term. 
(/)  A  polynomial  is  said  to  be  arranged  in  descending  powers 

of  one  of  its  letters  if  the  term  containing  the  highest  power  of 
that  letter  is  placed  first ;  if  the  next  lower  power  is  placed 
second ;  and  so  on. 

(g)  Any  factor  of  a  product  is  the  Coefficient  of  the  product 
of  the  remaining  factors.     If  one  factor  is  expressed  in  nu- 


THE    FUNDAMENTAL   OPERATIONS  7 

merals  and  the  remaining  factor  in  letters,  the  former  is  called 
the  Numerical  Coefficient  of  the  latter. 

(h)  A  Common  Factor  of  two  or  more  terms  is  a  factor  of  each 
of  them. 

(?')  Like  or  Similar  Terms  are  terms  that  are  alike  in  their 
literal  parts  ;  Unlike  or  Dissimilar  Terms  are  terms  which  are  not 
alike  in  their  literal  parts.  Terms  are  like  with  respect  to  one 
or  more  factors  if  they  have  these  factors  as  common  factors ; 
thus,  3  a(x  —  y)  and  4  b(x  —  y)  are  like  wTith  respect  to  (x  —  y). 

5.  Addition  and  Subtraction  of  Expressions. 
Rule.  —  To  add  two  or  more  like  terms : 

1.   Multiply  their  common  factor  by  the  sum  of  its  coefficients. 

Thus,  2  a(x  -  y)  +  3  b(x  -  y)  =  (2  a  +  3  b)  (x-y). 

This  rule  follows  from  the  distributive  law  of  multiplication, 
§1,/ 

Rule.  —  To  add  polynomials ; 

1.  Write  the  polynomials  with  like  terms  in  vertical  columns. 

2.  Add  the  columns  of  like  terms,  and  connect  the  results  by 
their  signs. 

This  rule  follows  from  the  commutative  and  associative  laws 
of  addition,  §  1,  b  and  c. 

Rule.  —  To  subtract  one  term  from  a  like  term  or  one  polynomial 
from  another : 

1.  Write  like  terms  in  vertical  columns. 

2.  Imagine  the  signs  of  the  terms  of  the  subtrahend  changed,  and 
add  the  resulting  terms  to  those  of  the  minuend. 

6.  Parentheses,  (  ),  Brackets,  [  ],  Braces,  j  J,  and  the  Vin- 
culum, ~,  are  symbols  of  grouping,  used  to  indicate  terms 
which  are  to  be  treated  as  parts  of  a  single  number  expression. 

Thus,  3a  -(2x  +  y  -  z)  means  that  2 x  +  y  -  z  is  to  be  subtracted 
from  3  a. 


8  ALGEBRA 

Rule.  —  To  remove  parentheses  preceded  by  a  plus  sign  : 
Rewrite  all  terms  which  are  within  the  parentheses  without 
changing  their  signs. 

Rule.  —  To  remove  parentheses  preceded  by  a  minus  sign : 
Rewrite  all  terms  which  are  within  the  parentheses  but  change 
their  signs  from  +  to  - ,  or  from  -  to  -f . 

Sometimes  terms  must  be  inclosed  within  parentheses. 

Rule.  —  1.  To  inclose  terms  within  parentheses  preceded  by  a  plus 
sign,  rewrite  the  terms  without  changing  their  signs.     . 

2.  To  inclose  terms  within  parentheses  preceded  by  a  minus  sign, 
rewrite  the  terms,  changing  their  signs  from  +  to  - ,  or  from  -  to  + . 

Thus,  r  +  s-t-r  +  (s-t)  =  r  —  (—  s  +  t). 

EXERCISE  2 

1.  Consider  the  monomial  5  ab2&{x  —  ?/). 

(a)  What  are  its  factors  ?  (b)  What  is  its  numerical  coeffi- 
cient ?  (c)  What  are  the  exponents  of  a,  b,  and  c  respectively  ? 
(d)  What  is  the  coefficient  of  (x  —  y)  ?  of  b2<?  ? 

2.  Arrange  3  xyz  —  2  x2y2  +  y4  +  x4  —  2  y?y  in  ascending 
powers  of  y. 

3.  (a)  2  mn(x  -f  y)  and  3  m2n(x  +  y)  are  like  with  respect 
to  what  common  factor  ?  (6)  What  is  the  coefficient  of  that 
factor  in  each  of  the  terms  ? 

4.  Add  7x  +  6y  —  9z  and  4=x  —  8y  +  5z. 

5.  Add  3x2  +  7y2-2xy,  9xy  -  5x2-  10y2,  and  8a;2- 
6  xy  —  4  y2. 

6.  Add  a-9-8a2  +  16a3,  5  +  15  a3  -  12  a  -2  a2,  and 
6a2-  10a3  +  11  a  -13. 

7.  Add   6(a  +  6)-6(c-d)   and   3(«  +  ?>)  +  8(c  -  d). 


THE   FUNDAMENTAL   OPERATIONS  9 

8.  Add    14(*  +  y)  —  1"0  +  z),    KV  +  z)  -  9(2  +  »)i    and 

—  3(x  +  y)—7(z  +  x). 

9.  Add  2  ax  +  3  &x  —  4  ex. 

10.  Add  5  mx2  -j-  2  nx?/  +J9?/2  and  tx2  —  rxy  +  qy2. 

11.  From  8 a; +  2//  —  7  z  subtract  8  x  —  2y  +  7  z. 

12.  Subtract  5  n3  -  9  -  14  n2  +  16  n  from  7n2  +  20n3-5 
+ 13  ». 

13.  Take  49  x2  -f  16  m2  -  56  mx  from  25  m2  +  36  x2  -  60  mx. 

14.  Subtract  —  5{a  +  b)  +  9(c  -  d)  from  7(a  +  6)  -  6  (c  —  a"). 

15.  From  3(x  +  ?/)2-2(x  +  2/)+5  take  (a?+y)2  +  3(a?  H-  y)  —  7. 

16.  What  expression  must  be  added  to  3  x2— x  -f  5  to  give  0? 

17.  By   how  much    does    2  m  —  4  m2  —  15  -f  17  m3    exceed 

—  9  +  6  m3  —  11  m  —  14  ra2  ? 

18.  From  the  sum  of  2x2  —  5xy  —  y2  and  7  x2  —  3  xy  +  4  ?/2, 
subtract  4  x2  —  6  xy  +  8  y2.  (Do  it  all  in  one  operation,  if 
possible.) 

19.  From  7  x—  5z  —  3y  subtract  the  sum  of  8  ?/  +  2  x  —  11  z 
and  6  z  —  12  y  +  4  x. 

20.  From  the  sum  of  7  x3  —  4  x2  +  6  x  and  3  x2  —  10  x  —  5,  take 
the  sum  of  -  5  x3  +  4  x  +  12  and  8  x3  -  11  x2  -  2. 

Remove  parentheses  and  combine  the  terms: 

21.  2x-3y  +  (5x-y)-(-$x  +  7y). 

22.  5  a  -(7  a  -[9  a  +  4]). 

23.  2x-(8?/  +  5x-J5x-?/D-(-92/  +  3x). 

24 .  8  a2  -  9  -  \  5  a2  -  ( 3  a  +  2)  J  +  j  6  a2  -  (4  a  -  7)  j . 

25.  5m-[7m-S-3m-(4w  +  9)|-56m-8|]. 


26.   25 -(-8 -[-34 -16 -47]). 


10  ALGEBRA 

Inclose  the  last  three  terms  of  the  following  in  parentheses 
preceded  by  a  minus  sign : 

27.  a2-4?>2  + 12  6  —  9.  29.    a2  +  b2  -  c2  +  d2. 

28.  4  x2  —  y2  —  2  yz  —  z2.  30.    n4  —  8n2-f  6  n ,  +  7. 

7.  Multiplication.  The  Law  of  Signs  for  Multiplication  is 
stated  in  paragraph  2,  (g). 

(a)  The  Law  of  Exponents.  The  exponent  of  any  number  in 
a  product  is  equal  to  its  exponent  in  the  multiplicand  plus  its 
exponent  in  the  multiplier. 

This  law  is  proved  for  certain  exponents  in  paragraph  115. 

Rule.  —  To  find  the  product  of  two  monomials  : 

1.  Find  the  product  of  their  numerical  coefficients,  using  the 
Law  of  Signs  for  Multiplication.     (§  2,  g.) 

2.  Multiply  this  result  by  the  product  of  the  literal  factors, 
using  the  Law  of  Exponents  for  Multiplication. 

This  rule  is  a  consequence  of  the  commutative  and  associa- 
tive laws  of  multiplication.     (§  1,  d  and  e.) 

Rule.  —  To  find  the  product  of  a  polynomial  and  a  monomial : 

1.  Multiply  each  term  of  the  polynomial  by  the  monomial. 

2.  Unite  the  results  with  their  signs. 

Rule.  —  To  find  the  product  of  a  polynomial  and  a  polynomial  : 

1.  Multiply  the  multiplicand  by  each  term  of  the  multiplier. 

2.  Add  the  partial  products. 

These  last  two  rules  are  consequences  of  the  distributive  law 
of  multiplication.     (§  1,  /.) 

It  is  desirable  to  arrange  both  multiplier  and  multiplicand 
according  to  the  same  order  of  powers  of  a  common  letter. 


THE   FUNDAMENTAL  OPERATIONS  11 

Example.     Multiply  x2  -  y2  4-  2  xy  by  y2  +  x2  -  2  a#.  t 

Solution  :  x2  +  2  xy  —  J8 

x2  -  2  xy  +  y2 
x4  +  2  x3y  —     x22/2 

-  2  x3^  -  4  x2y2  +  2  xy3 

x2y2  +  2  xy3  -  y4 


x4  —  4  x2y2  +  4  xy3  —  y4 

Note.     Multiplication  by  detached  coefficients  is  considered  in  §  256, 
and  may  be  studied  at  this  time,  if  desired. 

8.    One  of  the  useful  forms  of  multiplication  is  illustrated 
by  the  following  example. 

Example.     Find  the  product  of  2  x  —  3  y  and  a?  —  3xy  —  5 y2. 

Solution:  1.    (2  x  —  3  y)  (x2  -  3  xy  -  5  y2) 

2.  =  2  x(x2  -  3  xy  -  5  y2)  -  3  y(x2  -3xy-5y2) 

a  =  2  x3  -  6  x2y  -  10  xy2  -  3  x2y  +  9  xy2  4  15  y3 

4.  =  2  x3  -  9  x2y  -  xy2  +  15  */3. 

Note.    The  second  step  is  often  omitted. 

EXERCISE   3 

Multiply : 

1.  -  8  a4  by  7  ab\  3.    9(a  +  6) 2  by  3(a  +  6)3. 

2.  -5a^c  by  - 12  afc*  4.    13(x-y)  by  -2(z-2/)2. 

5.  10  a36  +  7  «64  by  -  6  ab\ 

6.  3  a2  -  2  a&  -  4  b2  by  -  4  a363. 

7.  m2  —  m  —  3  by—  2  m.  9.    a2  —  2  a&  4-  b2  by  a  —  6. 

8.  x2  —  2  «?/  +  4  ?/2  by  -  3  ay.      10.    c2  4-  2  ed  4-  d2  by  c  —  eZ. 

11.  a;2  —  2  xy  +  y2hy  x2  +  2  xy  4-  ?/2. 

12.  -  6  x  4-  2  a2  +  8  by  -  4  +  x2  4-  3  x. 

13.  a3  +  53  +  2  a62  4-  2  a26  by  b2  4-  a2  -  2  a6. 

14.    (a  4-  6  -2  c)2.        16.    (a-2  6)4.  18.    (^-m-^n* 

h:  {«-zyy.         17.  g*W;         19.  |*-|yv 


12  ALGEBRA 

Find  the  following  products  as  in  §  8 : 

20.  (2w2-4n-f7)(>  +  2). 

21.  (4  m2  +  9  n2  -  6  mn)(2  m  +  3  n). 

22.  (3a2-2a  +  4)(2a-l). 

23.  (5m2  +  3m-4)(6m-5). 

24.  (a2  +4  xy  -+  16  ?/2)(a  —  4  y). 

25.  (5r2-3rs  +  6s2)(2r-3.s). 

9.  Division.  The  Law  of  Signs  for  Division  is  stated  in  para- 
graph 2,  h. 

The  Law  of  Exponents.  The  exponent  of  any  number  in  a 
quotient  is  equal  to  its  exponent  in  the  dividend  minus  its 
exponent  in  the  divisor.  This  law  is  proved  for  certain  expo- 
nents in  paragraph  115. 

Rule.  —  To  divide  a  monomial  by  a  monomial : 

1.  Find  the  quotient  of  their  numerical  coefficients,  using  the 
Law  of  Signs  for  Division.     (§  2,  h.) 

2.  Multiply  the  result  by  the  product  of  their  literal  factors, 
using  the  Law  of  Exponents  for  Division. 

Rule.  —  To  divide  a  polynomial  by  a  monomial : 

1.  Divide  each  term  of  the  polynomial  by  the  monomial. 

2.  Unite  the  results  with  their  signs. 

Rule  . —  To  divide  a  polynomial  by  a  polynomial : 

1.  Arrange  the  dividend  and  the  divisor  in  either  ascending  or 
descending  powers  of  some  common  letter. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  and  write  the  result  as  the  first  term  of  the  quotient. 

3.  Multiply  the  whole  divisor  by  the  first  term  of  the  quotient ; 
write  the  product  under  the  dividend  and  subtract  it  from  the 
dividend. 


THE   FUNDAMENTAL   OPERATIONS  13 

4.  Consider  the  remainder  a  new  dividend,  and  repeat  steps,  i,  2, 
and  3. 

Example.     Divide  2  a4  +  8  a  -  a?  +  15  by  2  a2  -  3  a  +  5. 

<x2  +     a  -  1 

Solution  :    2  a2  -  3  a  +  5 1 2  a4  -     a3  +  8  a  +  15 

2  q*  -  3  a3  +  5  a2 

2  a3  -  5  a2  +  8  a  +  15 
2  a3  _  3  q-2  +  5  a 

-  2  d1  +  3  a  +  15 
-2a2  +  3a-    5 
+  20 
Quotient  =  a2  +  «  —  1  ;  remainder,  +  20. 

Note.     Division  by  detached  coefficients  is  considered  in  paragraph 
256,  and  may  be  studied  at  this  time,  if  desired. 

EXERCISE  4 

Divide : 

1.  6  xGy10  by  -  tfy10.  3.    9  (a  -  bf  by  3  (a  -  b)2. 

2.  _  45  a*b7  by  -  5  ab\  4.    32  (x+y)7  by  -4  (a+2/)3. 

5.  25  a8  -  15  a6  +  40  a4  by  5  a4. 

6.  -  24  msn2  +  33  ran7  by  -  3  mn2. 

7.  54  a465  -  60  a7b6  by  6  a&5. 

8.  -  22  xwy3  +  26  ^y  by  -  2  xAf. 

9.  6  a2  +  29  a  +  35  by  2  a  +  5. 
10.  30  x2  -  53  x  +  8  by  6  x  - 1. 

11.  a3-863by  a  -  2  &.  13.   x6  -  27  */3  by  x2  +  3  y. 

12.  a;4  +  ?/4  by  x  +  2/.  14.   243  r<5  +  1  by  3  w  + 1. 

15.  9  a4- 16  a2 +  8  a- 1  by  3  a2  +  4  a  -  1. 

16.  x2  —  y2  —  2  yz  —  z2  by  x  —  y  —  z. 

17.  2  or5  -  10  -  6  x2  +  xA  +  11  x  by  2  +  a2  -  a. 

18.  a4  -f-  a2?/2  +  ?/4  by  a2  —  a??/  +  y2. 

19.  3  n4  - 11  ns  -  25  n2  +  2  -  13  n  by  3  ?*2  +  4  n  + 1. 

20.  73  a; +  37  a3  -  35  +  20  a4  -  15  a;2  by  -5  +  4z2  +  9x. 


II.     SPECIAL   PRODUCTS   AND   FACTORING 

10.  Special  Products.  Of  the  special  products  studied  in  the 
first  course  in  algebra,  the  following  are  particularly  important : 

(a)  The  Product  of  the  Sum  and  the  Difference  of  (any)  two 
numbers  equals  the  difference  of  their  squares. 

(x  +  y)(x-y)  =  x2-y2. 

(b)  The  Square  of  a  Binomial  equals  the  square  of  its  first 

term,  plus  twice  the  product  of  its  two  terms,  plus  the  square 

of  its  second  term.    ,  .     _ 

(x  +  y)2  =  x2  +  2xy  +  y1. 

Note.  If  the  second  term  is  negative,  then  the  middle  term  of  the 
square  is  also  negative. 

Example.     (3  x -  5  yf  =  (3  xf  +  2(3  x)(-  5  y)  +  (-  5  y)2 
=  9  x2  -  30  xy  +  25  y2. 

(c)  .The  Product  of  Two  Binomials  Having  a  Common  Term. 

(x  4-  a)(x  +  ft)  =  x2+(fl  +  &)x  +  ab. 

Example,     (x  +  6)  (x  - 13)  =  x2  +  (6  - 13)  x  +  6  ( -  13) 

=  a,.2_7a,__78< 

(ri)  The  Product  of  Two  General  Binomials. 

(ax  +  b)(cx  +  d)  =  acx2  +  (be  +ad)x  +  bd. 

Example.     (3  a  -  4  6) (2  a  +  5  b)=  3  a  •  2  a  +(-  8  -f- 15) a& 
+  (-4  6)(5  6)=  6  a2  +  7  a&  -  20  62. 

Note.  It  is  highly  important  to  he  able  to  use  these  last  two  formulae 
readily,  — particularly  so  the  last,  as  it  includes  each  of  the  preceding  as 
a  special  case.  For  further  discussion  of  this  type,  if  necessary,  see  First 
Year  Algebra,  §  102. 

14 


SPECIAL   PRODUCTS    AND   FACTORING  15 

EXERCISE  5 

1.  To  what  type  do  the  following  belong  ?     Multiply  : 

(a)  (*  +  9X«-9).  (/)  (3m2-5?i)(3m2  +  5n). 

(b)  (m  +  8)(m-8).  (g)  (10 +  4)(10 -4). 

(c)  (k2-<12)(tf  +  12).  (A)  22  •  18. 

(d)  (2  x  -  3)(2  a?  +  3).  0')  33  .  27. 
(c)  (4r-7s)(4r-f-7s).  (j)  ^4  •  36. 

2.  To  what  type  do  the  following  belong  ?     Multiply  : 

(a)  (a  +  3)2.  (e)  (8m-3r)2.  (i)  (5a*/-2w)2. 

(6)  (m2  +  ll)2.  (/)  (r -  6  c)2.  (j)  (jp-2  q)\ 

(c)  (2*  +  5)2.  (g)  (*-8)«.  /2_a_c' 

(d)  (4z-3«))2.  (h)  (3t-l)\  K  }  \3b      d 
(/)  What  kind  of  an  expression  is  the  square  of  a  binomial  ? 

3.  To  what  type  do  the  following  belong  ?     Multiply : 

(a)  (aj  +  5)(»  +  6).      "  (g)  (t  +  6k)(t -10  k). 

(b)  (r  +  ll)(r  +  2).  (*)  (a  -  4  c)(a  + lie). 

(c)  (y  +  6p)(y  +  4i>).  (0  (r  +  5*)(r-9«). 

(d)  (*-5)(2-8).  (i)  (#  +  S4)(c?-12d), 

(e)  (w-3s)(w-7s).  (A;)  0?/  +  7  2)0*/  -  12  z). 
(/)  (a-5  6)(a  +  3  6).  (1)  (m3  -  3  r)(m3  +  11  r). 

(m)  What  is  the  sign  of  the  third  term  of  a  product  of  this 
type  when  the  signs  of  the  second  terms  of  the  binomials  are 
unlike  ? 

4.  To  what  type  do  the  following  belong  ?     Multiply : 
(a)  0  +  3)(2a  +  l).  (d)  (2aj-3)(2aj-l). 
(6)  (3a?  +  l)(2a>  +  l).  (•)  (3x-4y)(x-2y). 
(c)  (5  x  +  2)(*  +  3).                     (/)  (5  a;  -  3  y)(s  -  2  y). 


16  ALGEBRA 

(g)  (a  +  2d)(3a-&).  Q)  (llp  +  3q)(2p-3q) 

(h)  (c-3cZ)(3c  +  5d).  (m)  (2  xy  +  7)(3  ay  -  $). 
(i)  (3ffl-4w)(m  +  2)i).  («)  (10  rarc-3  «)(2  wm+3*), 

(j)  (7c-2d)(3c  +  4d).  (o)  (5*  +  6rs)(6*-7rs). 

(A;)  (2r*-5*)(3r*-i-2*).  (/>)  (12x2-5y)(3  x2  +  2y). 

Find  mentally  the  following  products  : 

5.  (5m -2n)2.  28.    (5  a8 -  6  ?/2)(5  aj3  +  y2). 

6.  (a  +  lly)(a+3y).  29.    (5a  +  i)(5a-i). 

7.  (x  +  12y)(x-2y).  30.    (4^-3^. 

8.  (2»-F3)(*+4).  31.    (3  y +7)0/ -5). 

9.  (a2-4  2/)(a2  +  4?/).       ,  32.    (2  -  3  ay) (5  +  2  xy). 

10.  (2cd-7)2.  33.    (x*y  +  6  z)(x*y  -  13  z). 

11.  (3 a2//-  4 2) (3 ^-4 2).  34.    (ay+5)(*p~4), 

12.  (2»i-5)(m  +  4).  35.    (a3  +  7)(a3-  11). 

13.  (2p2-7)(3i>2  +  5).  36.   (3a  +  5)(7*-8). 

14.  (r2-3s)0«2  +  7s).  37.    (l-9r)(l  +  8r). 

15.  (x  +  i)(x-  J).  38.    (3  c  + c/3)(2  c  -  d8). 

16.  (Jm  +  5p)(£m  —  5p).  39.    (5m-3]y)l 

17.  (a-i)(a-i).  40.    (12a-i)(9a-|). 

18.  (2a  +  3)(£a  +  l).  41.    (20  -  16  2) (3  +  2  z). 

19.  (3a2  +  4  6)(3a2-4  6).  42.    (a  +  16  ?/3)(a-  ?/3). 

20.  (y_10)(y  +  4).  43.    (2/  -  6  a2)(?/  +  a2). 

21.  (a  —  |)(a  +  f ).  44.    (4  r  +  s*)(4  r  —  5  st). 

22.  (l-6«)(3  +  4«).  45.    (6  m3 -|)2. 

23.  (2^-7^(3^-4^.  46.    (1  +23  z)(5  -z). 

24.  (3m_|)(3m_hi)>  47.    (5  a2- 4^(6  a2  -5  y). 

25.  (3t-7r)(2*  +  dr).  48.   (a(5-?/6)(a6  +  2/6). 

26.  (11«*-.1)(12*  +  1).  49.    (9a+2y)(3a-4y). 

27.  (z2-6)(z2  +  12).  50.    (12c  +  5d)(4c-3d). 


SPECIAL   PRODUCTS   AND   FACTORING  17 

FACTORING 

11.  A  monomial  is  said  to  be  Rational  and  Integral  when  it 
is  either  an  arithmetical  number  or  a  single  literal  number 
with  unity  for  its  exponent,  or  the  product  of  two  or  more  such 
numbers  ;  as,  3,  a,  or  2  a?bc2. 

12.  A  polynomial  is  said  to  be  rational  and  integral  when 
each  term  is  rational  and  integral ;  as,  2  a?b  —  3c  +  d2. 

13.  To  Factor  an  algebraic  expression  is  to  find  two  or  more 
expressions  which  will  produce  the  given  expression  when  they 
are  multiplied  together.  It  is  agreed  that  only  rational  and 
integral  factors  of  integral  expressions  will  be  considered. 

14.  A  number  which  has  no  facfors  except  itself  and  unity 
is  called  a  Prime  Number ;  as,  3,  a,  x  +  y. 

15.  Type  Forms.  Skill  in  factoring  depends  upon  ability  to 
recognize  the  type  forms.  The  following  forms  were  studied 
in  the  first  course  in  algebra : 

(a)  Removing  a  Monomial  Factor.     A  monomial  factor  of  an 

expression  is  a  number  which  will  exactly  divide  each  term  of 

the  expression.  , 

mx  +  my  +  mz  =  m(x  +  y  +  z). 

Example.     3afy  +  9afy  +  3a#  =  3  xy(x2  +  3x  +  l). 

(b)  The  Difference  of  Two  Squares  equals  the  product  of  the 
sum  and  the  difference  of  their  square  roots 

x2-y2  =  (x-y)(x  +  y). 

Example.       25  x*  —  yA  =  (5  a?  +  y2)(p  a3  -  y2)- 

(c)  A  Perfect  Square  Trinomial  is  one  which  has  two  terms 
that  are  perfect  squares  and  whose  remaining  term  is  twice  the 
product  of  their  square  roots. 

To  Find  the  Square  Root  of  a  perfect  square  trinomial,  ex- 
tract the  square  roots  of  the  two  perfect  square  terms,  and 
connecb  them  by  the  sign  of  the  remaining  term. 


18  ALGEBRA 

Example  1.  4  a;2-  12  x  +  25  is  not  a  perfect  square  ;  for 
V4^"2  =  2x,  V25  =  5,  and  2  .  2  x  •  5  =  20  x  and  not  12  a. 

Example  2.  4  a?  —  20  a#  +  25  ?/2  is  a  perfect  square  ;  its 
square  root  is  2  x  —  5.     Hence  4  #2  —  20  a??/  -f-  25  y2  =  (2  #  —  5  #)2. 

(d)  Trinomials  of  the  Form  x2+px  +  q  can  be  factored  when 
two  numbers  can  be  found  whose  product  is  q  and  whose  sum 
is  p. 

Example.  x2-S  x-  65  =  (x  +  5)  (x  -  13),  for  5(-13)  = 
- 65 ,  and  (+  5)  +  (-  13)  =  -  8. 

(e)  Trinomials  of  the  Form  ax2  +  bx  +  c,  if  factorable,  can  be 
factored  as  in  the  following  example : 

Example.     Factor  15  x2  -f-  17  x  —  4. 

Solution :  For  15 x2,  try  5  x  and  3  as,  thus  :  (5x  )(3a;  ) .  For 
—  4,  try  2  and  2,  with  unlike  signs,  arranging  the  signs  so  that  the  cross 
product  with  larger  absolute  value  shall  be  positive  ;  thus : 

(5  x  —  2  )(3  x  +  2).     Middle  term  +4x;  incorrect.     (See  §  10,  d.) 

For  —  4,  try  4  and  1,  arranging  the  signs  as  before  ;  thus : 

(5  x  +  4)  (3  x  —  1) .     Middle  term,  +  7  x  ;  incorrect. 
Try  (5x—  1)  (3  a  +  4).     Middle  term,  +  17  x  ;  correct. 

Note.  This  method  of  factoring  the  general  trinomial  is  the  most  use- 
ful. While  at  first  it  may  seem  difficult,  it  is  easily  mastered.  It  applies 
also  to  each  of  the  preceding  forms.     (See  First  Year  Algebra,  §  103.) 

(/)    The  Sum  of  Two  Cubes,     x3  +  f  =  (x  +  y)(x2  -  xy  +  y2). 

Example,     m6  +  27  f  =  (m2)3  +  (3  y)z 

=  (m2  +  3  ?/)(ra4  -  3  mhj  +  9  y2). 

(a)  The  Difference  of  Two  Cubes,   x3  —  #3=(x  —  y){x2  +  xy  +  y2). 

Example.     216  a9  -  b*  =  (6  a3)~  -  53 

=  (6a3-  6)(36  a6  +  6  a36  +  62). 


SPECIAL   PRODUCTS   AND   FACTORING  19 

16.  Complete  Factoring.  First  remove  any  monomial  factor 
present  in  the  expression ;  then  factor  the  resulting  expres- 
sions by  any  of  the  preceding  type  forms  which  apply  until 
prime  factors  have  been  obtained. 

Example  1.     3  of  -  3  tf  =  3(>6  -  f)  =  3(ar3  +  y3)  (a?  -  yz) 

=  3{x  -f-  y)  (x2  —  xy  +  y2)(x  —  y)  (x2  +  xy  +  y2). 

Example  2.     24  ax2  +  22  axy  -  10  ay2 

=  2  a(12  x2  +  11  xy  -  5  y2)  =  2  a(4  x  +  5  y)(3  a;  -  y). 

Note.  Advanced  Topics  in  Special  Products  and  Factoring  are  dis- 
cussed in  Chapter  IX.  This  chapter,  in  whole  or  in  part,  as  far  as  para- 
graph 95,  may  be  studied  at  this  time  if  desired. 

EXERCISE  6 

1.  Remove  the  monomial  factor  : 

(a)  2  ax  —  6  ay  +  7  az.  (/)  m (a  —  2)  4-  n(a  —2). 

(6)  IS  mhi- 15  mn2.  (g)  a(x>  -  3)  -5(x2  -3). 

(c)  4  r1  -  8  r2 -  4  r.  (h)  5a(x-y)-3  b(x  - y). 

(d)  7n7  -7  n.  (i)  a(m2  -  2)  -  3(m2  -  2). 

(e)  a6  -  5  a5  -  2  a4  +  3  a3.  (?)  z2(a  -  6)  +  x\c  -  d). 

2.  To  what  type  do  the  following  belong  ?     Factor  : 

(a)  x2-36.  (/)  A^-iV^- 

(6)  4  m2 -81.  (g)  l-64m2n2. 

(c)  9  «2  -  4  2/2.  (h)  36  a2  -  121  y2. 

(d)  25  m4  -  1.  (i)  1-81  a262c2. 

(e)  x»-f.  (j)  226- f. 

3.  Supply  the  missing  term  so  as  to  make  perfect  squares  of: 
(a)  o2+(     )+64.  '  (c)    a2-(     )  +  9  62. 

(6)   m2-(     )  +  144.  (d)  4x2  +  (     )  +  25. 


20  ALGEBRA 

(e)    9m2*i2-(     )+4.  (h)  m4-14m2  +  (     ). 

(/)   ^  +  8«  +  (     ).  (0  eP-8<f+(    ). 

(<?)    y-122/2  +  (     ).  (J)  9z2  +  6z+(     )• 

4.  To   what   type    do    the    following    belong?     Factor,   i 
possible : 

(a)  4a2-20a  +  25.  (/)  25m6-20m3  +  4. 

(b)  9c2  +  6cd  +  d2.  (gf)  a2*/2 -14  ay +  49. 

(c)  16r2  +  12r  +  9.  (A)  81  m2  +  90  mn  +  100  n\ 

(d)  100ajy-20ay  +  l.  (t)  36  a2  - 132  a&  +  121  b\ 

(e)  49*4-42*2  +  9.  (j)  4  a6-  4a36c2+62c4. 

5.  To  what  type  do  the  following  belong  ?     Factor : 

(a)  c2  +  9c  +  20.  (0  a4  + 13  a2 - 42. 

(b)  m2- 10m  +  21.  (j)  t2  +  3*- 28. 

(c)  r4- 11^ +  30.  (fc)  W2_4^_45V2 

(d)  x2  +  6x-27.  (I)  a«-Cia3b-55b\ 

(e)  a?tf-2xy-35.  (m)  a*  +  l4%-i5$» 
(/)  a6  +  a3 -110.  (n)  1  +  5a -14a2. 
(g)   a2  +  12a#  +  20;r.  (o)  l-9a-36z2. 
(ft)  a262  - 17  abc-  60  c2.  (p)  l-13n-68n2. 

6.  To  what  type  do  the  following  belong  ?     Factor  : 
(a)   6r2-7r  +  2.  (fc)  35*  +  .* -6  iff; 

(6)   3a?  +  8a;  +  4.  (i)  9x2-14^/-82/2. 

(c)  6aj*-*-2.  (j)  12  zc2  -  35  w  -  3. 

(d)  9c2  +  15c  +  4.  (A:)  6 -a- 15a2. 

(e)  6x2-7a-10.  (0  5  +  92/-18/. 

(/)  14  f  +  13  2/2-  12  z\  (m)  24  ;r2  -  17  nx  +  3  w2. 

(gr)   8w2-2«;i;-21v2.  (w)  28.r2-a-2. 


n°. 


SPECIAL   PRODUCTS   AND   FACTORING  21 

7.  To  what  type  do  the  following  belong  ?     Factor  : 

(a)  a? -8.  (c)    x«-m\  (e)    27  w3-  8. 

(b)  c3-64d3.  (d)  8a3-l.  (/)  125r6-216ol. 

8.  To  what  type  do  the  following  belong ?     Factor: 

(a)  ^  +  8  a3,  (c)   a3  +  i.  (e)    27  a6  + 125  y9. 

(6)  m6  +  125w3.  (d)  c6  +  d12.  (/)  64*»»  +  ^ 

Factor  the  following : 

9.  9«2-64?/4.  30.  4a2-3a-7. 

10.  6ax  —  10ay  +  2az.  31.  9a2  +  12a-32. 

11.  a2-f2z-35.  32.  a2-fl0ab  +  25&2. 

12.  18 A;2- 32 P,  33.  xl2-y12. 

13.  2/6  +  G?/3-27.  34.  6a2  +  7ax  +  2a2. 

14.  c\Z2  +  5  cd  -  66.  35.  25  x2  -  25  mx-Qm\ 

15.  62oay-Jg.  36.  3x*- 12. 

16.  3  ccfy2  -  9  ccty  -  30  cd.  37.  9  m2  -  42  m*  + 49  *2. 

17.  4  ax2  -25  ay4.  38.  10  x2  -  39  a  + 14. 

18.  3?/3  +  24.  39.  12.x2  +  lla  +  2. 

19.  4x2 -27 x -{-4:5.  40.  36 a2 +  12 a; -35. 

20.  6x2  +  7x-3.  41.  ^-82/3. 

21.  &z2-l.  42.  2  am2  -50  a. 

22.  Wxzy-5x2y2-5xtf.  43.  72  +  7  a  -  49  a2. 

23.  mhi2  +  7  mw  -  30.  44.  31  a2  +  23xy-Sy2. 

24.  a2-3a-2/-70?/2.  45.  24a2  +  26a-5. 

25.  mx2  +  7  ma  —  44 m.  46.  l-3a?/  —  108  a2?/2. 

26.  a3-  3  x2  -108  a.  47.  x2  -  14  mx  +  40  m2. 

27.  a8-?/8.  48.  2fr  +  10a&-28a26. 

28.  xA-5x2y-24y2.  *        49.  c3  +  27d3. 

29.  8?*2+  18^-5.  50.  Zx?y-21xtf. 


22  ALGEBRA 

51.  3m10-48rn5-240r.  56.  1-64  a6. 

52.  36  x4-  121  if.  57.  x*y-16xy\ 

53.  mhn  +  m3n4  —  ??iw6.  58.  a6  —  5  a5  +  2  a4  —  3  a2. 

54.  5#y  +  70a*/  +  245.  59.  _  16  a2  +  2±ax-  9  x2. 

55.  20m2  +  9w^-18w2.  60.  20  aV  -  23  aa;  4- 6. 

17.  The  Degree  of  a  Rational  and  Integral  Monomial  (§  11)  is 

the  sum  of  the  exponents  of  its  literal  factors. 

Thus,  a4bc3  is  of  the  eighth  degree. 

18.  The  degree  of  a  rational  and  integral  polynomial  (§  12)  is 
the  degree  of  its  term  of  highest  degree. 

Thus,  2  a2b  -  3  c  -f  d2  is  of  the  third  degree. 

19.  The  Highest  Common  Factor  (H.  C.  F.)  of  two  or  more 
rational  and  integral  expressions  is  the  expression  of  highest 
degree,  with  greatest  numerical  coefficient,  which  will  divide 
each  of  them  without  a  remainder. 

Rule.  —  To  find  the  H.  C.  F.  of  two  or  more  expressions : 

1.  Find  the  prime  factors  of  each  expression. 

2.  Select  the  factors  common  to  all  the  given  expressions,  and 
give  each  the  lowest  exponent  it  has  in  any  of  the  expressions. 

3.  Form  the  product  of  the  common  factors  selected  in  step  2. 

Example.     36  (r  +  s)\r  -  sf  =  2  .  2  .  3  •  3  (r  +  s)\r  -  sf 

=  22  •  32  •  (r  +  sf(r  -  sf. 

30 (r  +  s)(r -  s)4  =  2  .  3  •  5(r  +  «)(r -  s)\ 

.-.  the  H.  C.  F.  =  2  .  3  (r  -f  «)(r  -  sf  =  6 (r  +  s)(r  -  s)\ 

20.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
rational  and  integral  expressions  is  the  expression  of  lowest 
degree,  with  least  numerical  coefficient,  which  can  be  exactly 
divided  by  each  of  them. 


SPECIAL   PRODUCTS   AND   FACTORING  23 

Rule.  —  To  find  the  L.  C.  M.  of  two  or  more  expressions  : 

1.  Find  the  prime  factors  of  each  of  the  expressions. 

2.  Select  all  of  the  different  prime  factors  and  give  to  each  the 
highest  exponent  with  which  it  occurs  in  any  of  the  expressions. 

3.  Form  the  product  of  all  of  the  factors  selected  in  step  2. 

Example.     40  mzn2  (m  —  n)\m  -+-  n)3 

=  23  •  5  •  m3n2(m  —  n)\m  +  nf. 
10  mw3  (m  —  ri)\m  -f-  n)  =  2  •  5  •  mnz  (m  —  n)3(m  -f-  n). 
.-.  the  L.  C.  M.  =  23  •  5  •  mhi*  (m  -  n)\m  +  nf 
=  40  ra3n3  (m  —  n)3(ra  +  n)3. 

EXERCISE  7 
Find  the  H.  C.  F.  and  also  the  L.  C.  M.  of: 

1.  18  and  45.  3.    28  mhi  and  2  rn?n\ 

2.  24  *y  and  30  xy\  4.   35  a*b  and  15  a4. 

5.  (r  +  s)2(r  —  s)  and  3r(r+  s)(r  —  s)2. 

6.  a2  -  4  62  and  a2  +  4  a&  -f  4  62. 

7.  5  afy(a  +  3)  and  10  a^2(a  +  3). 

8.  a2  +  2  a  -  3  and  a3  -  1. 

9.  9  —  x2  and  #2  —  x  —  6. 

10.  a2 +  2 a; -24  and  x2-x-42. 

11.  8a3-63and2a2  +  5a6-352. 

12.  3arJ-32/3and2£2  +  4a?/-62/2. 

13.  6a2  +  5ad-4d2and  6a2  +  ad-2d2. 

14.  x2  —  9  2/2,  a;2  —  a-?/  —  6  #2,  and  a;2  —  6  #?/  -f  9  ?/2. 

15.  2  a  + 10  b,  4  a2  +  2  a&,  and  a2  + 10  ab  +  25  62. 


III.     FRACTIONS 

21.  A  Fraction  indicates  the  quotient  of  the  Numerator 
divided  by  the  Denominator.  The  numerator  and  denomi- 
nator are  called  the  Terms  of  the  fraction. 

22.  Signs  of  a  Fraction.     By  the  rules  of  signs  for  division, 

±^  =  -±^  =  -H3-  also,  +^  =  ^- 
+  6  -b  -f  b '  +  b      -b 

Rule.  —  1.  If  the  sign  of  one  term  of  a  fraction  is  changed,  the 
sign  of  the  fraction  itself  must  be  changed. 

2.  If  the  signs  of  bcth  terms  of  a  fraction  are  changed,  the  sign 
of  the  fraction  itself  must  not  be  changed. 

tj,  2a-3         3-2z         2a-3     3-2z 

Example. =  — =  — —  = —  • 

4a;  — 5  4sc  —  a  5  —  4x      o  — 4sc 

23.  To  Reduce  a  Fraction  to  its  Lowest  Terms : 
Rule.  —  1.    Find  the  prime  factors  of  both  terms. 

2.   Divide  both  terms  by  all  of  their  common  factors. 

ExAMPLE.     a'  +  3a6-28yas(a-46)(q  +  76) 
16  b2-  a2  (4  6  -  a)(4  6  +  a) 


1 
lfl~***f){a  +  7&)=      a -{-lb 
&^¥&){a  +  4  6)  a  +  4  b 

1 


(See  §  22.) 


24.    An  Integral  Expression  is  an  expression  which  has  no 
fractional  literal  part ;  as,  a2  —  2  ab  +  b2,  or  f  ab2. 

A  Mixed  Expression  is  an  expression  which  has  both  integral 

and  fractional  literal  parts ;  as,  a  H 

c 
24 


FRACTIONS  25 

Rule.  —  A  fraction  may  be  i  educed  to  an  integral  or  a  mixed  ex- 
pression by  performing  the  indicated  division. 

^  x1 -'2x11  +  2  1?  ,      y2 

Example.     2-2- — •—  =  x  —  y-\ — 2 

x—y  x—y 

The  quotient  and  the  remainder  were  obtained  in  this  example  by  per- 
forming the  division  as  in  paragraph  9. 

25.  The  Lowest  Common  Denominator  (L.  C.  D.)  of  two  or 
more  fractions  is  the  lowest  common  multiple  (§  20)  of  their 
denominators. 

Rule.  — To  reduce  fractions  to  their  lowest  common  denominator : 

1.  For  the  L.  C.  D.,  find  the  L.  C.  M.  (§  20)  of  their  denominators. 

2.  For  each  fraction,  divide  the  L.  C.  D.  by  the  given  denomi- 
nator and  multiply  both  numerator  and  denominator  by  the 
quotient. 

Example.  Reduce  the  following  fractions  to  respectively 
equivalent  fractions  having  their  lowest  common  denominator: 

4a     and  Sa 


Solution  :    1. 


4 

4a 


a- 
3  a 


a1 

—  5  a 
4a 

+  6 

(a 

+  2)(a 
3a 

-2) 

a--5a+6      (a_3)(a_2) 

2.  The  L.  C.  D.  is  (a  -  2)  (a  -  3)  (a  +  2). 

3.  For  ^—  ■  the  L.  C.  D.  -  (a  -  2)(a  +  2)  =  a  -  3. 

.      4a    _  4a(a-3) 

"'a2-4      (a-2)(a-3)(a  +  2)  ' 

4.  For     ,     Sra :theL.C.D.  -(a  -  2)(a  -  3)=  a  +2. 

a2-5a  +  G  v  n  J 

■  3  a  _  3a(a  +  2) 

"  «2-5rt  +  li      (a_2)(a-3)(a  +  2)' 

In  each  case,  the  resulting  fraction  has  the  same  value  as  the  original 
fraction  ;  it  is  equivalent  to  the  original  fraction.  The  form  only  of  the 
fraction  has  been  changed. 


26  ALGEBRA 


EXERCISE  8 

Write  each  of  the  following  fractions  in  three  other  ways 
without  changing  the  value  of  the  fraction  : 

-        8  2x-7       n     a-b  M  6x-5 


2-x  5-x  d-c  (a?-3)(a;+4) 

5.  Prove  that  the  three  fractions  obtained  in  Example  1  are 
equivalent  to  the  given  fraction  by  finding  the  values  of  all 
four  fractions  when  3  is  substituted  for  x. 

6.  State  the  fundamental  principle  employed  in  reducing  a 
fraction  to  its  lowest  terms.  Is  the  value  of  the  fraction 
changed  ? 

Reduce  to  their  lowest  terms  : 

-     120  _     63aty*        a     90  mV      ,A     120  aW° 

390  84a?y  36  mW  75  a&V 

n     ff2-9a;  +  18  a3  +  &3 


x^-j-x-12 

a?  +  11  ab  +  28  62 
a2  +  14  ab  +  49  62 

a-2 -25 

12.     .  '  , .      '    :-:- •  15. 


a2  _  2  a&  -  3  62 

q2  +  a  _  12 
3  a2 -13  a +  12* 

9,2-49?/2    , 
a?  -  11  *  +  30  28  xy*  -  12  afy 

17.  If  the  value  of  a  fraction  is  desired  for  specified  values 
of  the  letters  involved,  should  the  substitution  be  made  in  the 
original  fraction  or  in  the  simplified  fraction  ? 

Find  the  value  of  the  fraction  in  Example  12  when  a  =  12 
and  b  =  5. 

Reduce  to  mixed  or  integral  expressions : 

lg     15  m2  +  12  m  +  4  35  x*  +  9  x  +  3 

3m  5a+2 

9  a2 +  2  30a5-5a4  +  15a2-7 

'    3»-l"  5  a2 


FRACTIONS  27 

22.    ^-y3-  23.    3«3  +  8<*2-4. 

x  —  ?/  a2  -f  2  a  —  3 

24.    Prove  that  the  original  fraction  and  the  mixed  expression 
obtained  from  it  in  Example  18  are  equivalent  for  m  —  2. 

Note  that  this  ise  means  of  checking  the  solution. 
Reduce  to  their  lowest  common  denominator : 


25. 

3      4       1 

8"'     5>     3" 

28. 

a  —  a  u    a  u  —  u 

3a26   '     2a62  ' 

26. 

1  a?b    3¥c   2c2a 
6    '    10  '     15  * 

29. 

4  a2             2 

a2  -  9'  3  a2  -  9  a 

27. 

5           4           6 
2  mhi    3  mn?  5  mn 

30. 

5                   3n 

m2  —  4  m  -h  4'  m2  —  4 

31. 

3  a                 2 

a3  +  27'  a2 -a-  12" 

32. 

1             3  mn           2  m2? 

i2 

ft)2' 

m  —  n'  2(m  —  n)2'  3(m  — 

33. 

2           4           6 

x  +  2'  a  -  2'  a2  -  3' 

34. 

a  +  3  6                  a  — 

36 

a  +  46 

a2-7«&  +  12&2'  a2_a& 

-12  62'  a2-9  62' 

IK 

2x  +  3                 x  +  2 

a?  —  5 

a-2  +  3  a.  _  io7  2  a2  +  7  a  -  157  2  a2  -  7  x  +  6 

26.    Addition  and  Subtraction  of  Fractions. 

Rule.  —  1.  Reduce  the  fractions,  if  necessary,  to  respectively- 
equivalent  fractions  having  their  lowest  common  denominator. 
(S  25). 

2.  For  the  numerator  of  the  result,  combine  the  numerators  of 
the  resulting  fractions,  in  parentheses,  preceding  each  by  the  sign 
of  its  fraction. 


28  ALGEBRA 

3.  For  the  denominator  of  the  result,  write  the  L.  C.  D- 

4.  Simplify  the  numerator  and  reduce  the  fraction  to  lowest 
terms. 

(§22) 


Example. 

la                1                a 

a2  —  x2      x2  —  a3      a2  —  x2      a3  —  x3 

1 

a 

(a  —  x)(a 

+  x)      (a  —  x)  (a2  +  ax  +  x2) 

w 

!  +  ax  +  x2)                                    a(a  +  x) 

(a  —  x) (a  +  x)  (a2  +  ax  +  a2)       (a  —  x) (a  +  x)  (a2  +  ax  +  x2) 

_      (a2  +  ax  +  x2)  -  a(a  +  x) x2 

(a  —  x)(a  +  x)(a2  +  ax  4-  x2)       (a  — x)(a  +  x)(a2+  ax  +  x2)' 


EXERCISE  9 
Perform  the  indicated  operations : 

4  a;  +  7       6  a;  —  5  1 1 

10  15  '    I6x2-8x  +  l      16z2-l 

3  a  -  8      4  a  -  9  Qa-1      a  4- 1   ,  4  a2  + 1 

/£. — .  O. 


9                12 

3. 

x—y     y—2z.z—Sx 

xy          2yz          3  zx 

\ 

m              2 

m  —  2      ra  4-  2 

5. 

a+3_a-3 
a  —  3      a  +  3    . 

9. 


10. 


11. 


6.    «  +  3y__a?-3y 


a  +  1 

a_l       a2_l  ' 

5  a? 

4a24-3a-l 

a-3 

a;2  +  x  _  12  * 

3^  +  2 
3a-2 

9  a2 +  4 
9  x2  -  4' 

a;2 

2a3 

x2  —  xy  -\-  y2      x?  +  y* 

1 

4a -1 

a_32/      a  +  Sy  a(2a-l)       (2  a- l)3 


m  —  1      m  4- 1      m  —  6 

m  —  2      ra  4-  2      4  —  ra5 

14. * +         » 


a2H-4a&  +  462     462-a2 


15. 


FRACTIONS  29 

1 


2a;2  +  5a;  +  3      4z2  +  8a  +  3 


n_        a  —  n        3(i-4m.3«-5m 

Id. -p 

2a-\-2n     3  a  -f  3  n     6  a  -j-  6  w 
a  a 


17. 


a2  +  4a-60      a2-4a-12 


18.    -J--^-  +  .      »- 6 


n  +  4      1  —  n     w2  +  3w-4 

19.     1 + I + I 

(a-b)(a  -  c)      (6  -  c)(6-  a)      (c  -  a)(c  -  b) 

20.  *-3  +      ^  +  27 


a2  + 3a  +  9 

27.    Multiplication  and  Division  of  Fractions. 

Rule.  —  To  find  the  product  of  two  or  more  fractions : 

1.  Find  the  prime  factors  of  the  numerators  and  denominators 
of  the  fractions. 

2.  Divide  out  (cancel)  factors  common  to  a  numerator  and  a 
denominator. 

3.  Multiply  the  remaining  factors  of  the  numerators  for  the 
numerator  of  the  product,  and  of  the  denominators  for  the  denomi- 
nator of  the  product. 

Rule.  —  To  divide  one  fraction  by  another : 

1.  Invert  the  divisor  fraction. 

2.  Multiply  the  dividend  by  the  inverted  divisor. 

Note  1.  In  problems  involving  both  multiplication  and  division,  per- 
form the  operations  in  order  from  left  to  right.     (§  3.) 

Note  2.  Integral  or  mixed  expressions  should  first  be  reduced  to 
fractional  form. 


30  ALGEBRA 

\vl      vx      xlJ  \        x  —  vj\v      X 

__  fx2  —  2vx  +  v2\  (x  —  v  +  2v\  _^_  (x2-v2\ 

\  X2^2  J\       X—V       J       \      XV      J 

_  (x  —  v)2     /#-f-?A  xv  _  1 

x2v2         \x  —  vj      (x  —  v)  (x  +  v)      XV 

Complex  Fractions  are  a  special  case  of  division  of  fractions. 
x2-y2      \    2    J     \     6     J        2         (x- 


x+y 
2 

x2  —  y2 

fx+y 

6 

3 

■y){?+y) 


x-y 

EXERCISE  10 
Perform  the  following  multiplications : 


i.  *: 

42 

36                                  d     n2-36           In2 
35                                           4ti2       n2  +  n-4:2 

2      6  am 
27  M 

■2      Qa    5                             a2 -2  a- 35     4  a3- 9  a 

—  •  36  w5.                       5. •  - 

i5                                          2  a3 -3  a2        7  (a -7) 

3     5a' 

'    Sb2 

963      7  c4,                             5z  +  2          ,        Q) 
10c3    6a4                    '    2a2  +  a-10     S         ;* 

7 

4  m2  4-8ra  +  3     6  m2 -9  m 

2  ra2  -  5  m  +  3       4  m2  -  1 

g 

16a-4     20a  +  5     x2-\-2x  +  l 

ox  —  5       6  #  -f  6        16  #2  —  1 

9 

cc3  +  8  y3    x  —  2y     x2  4-  2  xy  4-  4  ?/2 

a^  —  8  #3     a;  +  2  ?/     a;2  —  2  xy  +  4  ?/2 

10. 

2  n2  -  n  -  3      w2  +  4w  +  4     n2  -  w  -  2 
n4-8w2  +  16          n2  +  ri          2w2-3n 

FRACTIONS  31 


Perform  the  following  divisions  : 

11     4«2-25  .  /o„      kn  12    a2-a6-262  .  a-26 

a*-3xy  .  ar2-10  3;y  +  21?/2 
/jj  _  ^3  a?2  -j-  icy  H~  2/2 

14      8n3  + 1    .  4n2-2n  +  l, 
2?i2  +  4n  *    w2  +  4n  +  4  ' 

2  a2  _  a&  _  3  &2         3  a2  +  a6  _  2  b2 


15. 


9  a2  -  25  62         9  a2  -  30  ab  +  25  62 


Perform  the  indicated  operations  : 

6a2-«-2         12a2-5a-2  4  a2  -  9 


16 


17. 


4a2-16a  +  15     8a2-18a-5     4a2+6a  +  2 

4  a2  _ 4 a  + 1     2a2  +  a  .  a2-2a 
4  a2  -  1        '  8  a3  -  1        a2  -  4  ' 


18  2x2-5xy-3y2  .  /2a;2  -  7  xy  -  4  y2  .  a2  -  4  an/ +  4  y  "y 

x2  —  xy  —  2y2      '  \x2 —  Sxy —  4=y2         x2  —  xy  —  6  y2 

19  /q  +  2    ■  2  V  a 3_\ 

V  «         a-3J\a-2     a  +  SJ 

20.    f2_^  +  4x-2n      /,jl      ^ 

V  x*  +  2x-8J     \x-2     x  + 


V  x  +  4  J      \  x  +  4  J 

„.  H.  2,  _J£  24.      « 

(X      —  -L  ~r 

x     y  9  a  m 


32  ALGEBRA 

28.     1  — 
25. 


26. 


27  a2 
b2 

_6 
a 

9a      Q      6 
o               a 

1              1 

1-a 

i-f  as 

1 

1 

tf2        1  +  X2 


K)K) 


29. 


1  +  tt 

2 


5a-      *•-* 


1_2aL-f_5 
3a-2 


X2_tfy_±tf 


27.    > 1Z^ ZJL.  30. 


a  +  2  +  -  x-f--^— 

x  %-\-y 


IV.     SIMPLE   EQUATIONS 

28.  An  Equation  expresses  the  equality  of  two  numbers. 
The  two  parts  (numbers)  are  called  respectively  the  Left 
Member  and  the  Right  Member  of  the  equation. 

29.  An  Identity  or  Identical  Equation  is  an  equation  in  which 
the  two  members  may  be  made  to  take  exactly  the  same  form 
by  performing  the  indicated  operations.     As, 

(a)  10  +  2  =  15  +  2-4-11.         (b)  3x(a-  b)=3ax-3bx. 

(a)  is  a  numerical  identity ;  (b)  becomes  a  numerical  identity 
for  any  values  of  the  literal  numbers. 

30.  An  equation  is  said  to  be  satisfied  by  a  set  of  values  of 
the  letters  involved  in  it  when  it  becomes  a  numerical  identity 
if  these  values  are  substituted  for  the  letters. 

31.  A  Conditional  Equation  is  an  equation  involving  one  or 
more  literal  numbers  which  is  not  satisfied  by  all  values  of  the 
literal  numbers. 

Thus,  Sx  —  4  =  aj  +  6  is  an  equality  only  when  x  =  5. 

The  word  equation  usually  refers  to  a  conditional  equation. 

32.  If  an  equation  involves  two  or  more  literal  numbers,  one 
or  more  of  the  literal  numbers  may  be  regarded  as  unknowns 
and  the  remaining  ones  as  known  numbers. 

Thus,  in  ax  +  by  =  c,  x  and  y  may  be  unknowns  and  a,  6,  and  c  may  be 
known  numbers. 

33.  If  an  equation  has  only  one  unknown  number,  any  value 
of  the  unknown  number  which  satisfies  the  equation  is  called 
a  Root  of  the  equation. 

33 


34        '  ALGEBRA 

(a)  x2  -f-  6  =  5  x  has  the  roots  2  and  3. 

When  x  =  2,  22  +  6  =  5  •  2,  for  each  equals  10.     ' 
When  a;  =  3,  32  +  6  =  5  •  3,  f or  each  equals  15. 

(b)  2ax  —  a  —  x  —  \  has  the  root  \. 

When  x  —  i,  2  a  •  \  —  a  =  |  —  j-  for  each  equals  0. 

34.  To  Solve  an  equation  is  to  find  its  root  or  roots. 
Example.   1.    Solve  the  equation      2  x  —  3  =  x~*~    • 

2.  Multiply  both  members  by  2.  4  #  —  6  =  x  +  6. 

3.  Subtract  #  from  both  members.  3  x  —  6  =  6. 

4.  Add  6  to  both  members.  3  x  =  12. 

5.  Divide  both  members  by  3.  #  =  4. 

The  problem  is  to  find  the  number  or  numbers  which  satisfy 
the  given  equation.  To  make  certain  that  4  satisfies  the  given 
equation  substitute  4  for  x  in  the  equation. 

Does2.4-3  =  t±5?     Does  8-3  =  —  ?     Yes. 

2  2 

This  test  is  called  checking  by  substitution. 

35.  Axioms  Used  in  Solving  an  Equation.  In  the  solution 
of  the  equation  in  paragraph  34,  the  following  axioms  were 
assumed : 

(a)  The  same  number  may  be  added  to  both  members  of  an 
equation  without  destroying  the  equality. 

(b)  The  same  number  may  be  subtracted  from  both  members  of 
an  equation  without  destroying  the  equality. 

(c)  Both  members  of  an  equation  may  be  multiplied  by  the 
same  number  without  destroying  the  equality. 

(d)  Both  members  of  an  equation  may  be  divided  by  the  same 
number  without  destroying  the  equality. 

As  long  as  the  number  used  as  an  addend,  subtrahend,  multi- 
plier, or  divisor  in  connection  with  these  axioms  is  an  arith- 


SIMPLE   EQUATIONS  35 

metical  number  (or  constant  other  than  zero),  the  equation 
obtained  by  applying  any  one  of  the  axioms  has  exactly  the 
same  roots  as  the  given  equation,  provided  the  given  equation 
has  a  root  at  all. 

36.  In  this  text,  symbols  A,  S,  M,  and  D  are  used  to  abbre- 
viate the  explanations  of  solutions  of  equations.     Thus  : 

A5:  means  "  add  5  to  both  members  of  the  previous  equation." 

S_3n :  means  "  subtract  —  3  n  from  both  members  of  the 
previous  equation." 

M_x:  means  "multiply  both  members  of  the  previous  equa- 
tion by  —  1." 

D4:  means  "divide  both  members  of  the  previous  equation 
by  4." 

Example.     Solve  the  equation  ^— ? -  2x-3  _  8_-3z     1 
4  2^  3  5 

1.  M30:         15(a;- 2)- 10(2  3 -3)  =6(8 -3s) +  30. 

2.  .-.  15  x  -  30  -  20  x  +  30  =  48  -  18  x  +  30. 

3.  .-.  -^5  x  =  78  -  18  x. 

4.  Ai8x:  13ic  =  78. 

5.  D13:  x  =  6. 

Check:     Does  «-2  _  12-3  =  8_^18 

2  3  5 

Does  2-3=-2  +  l?    Yes. 

EXERCISE   11 

Determine  by  substitution  which  of  the  numbers  1,  2,  —5, 
and  \  are  roots  of  each  of  the  equations. 


1.  4r+-5  =  6+-3r. 

2.  16s-l=4s  +  5. 

3.  w;2-3m?=-2.                         5. 

4,-t        12 
1-t     3-t 

2a2-3z+l  =  0. 

Solve  the  following  equations  : 

6.  10  x-  3  =  4  +  3  x,                    8. 

7.  21  y-  23  =  51  -16  y.             9. 

3(2«-l)  =  8(aj-l). 

5/_i_4/  —  41 

36  ALGEBRA 

10.  2a  — |a  +  |a  =  10. 

11.  fz  =  Jtf-fa;  +  -V3. 

12.  4(2v-7)+ 5  =  5(v-3)  +  16. 

13.  a- 2(4  -  7  a;)  =  4  aj- 9(2 -3  a). 

14.  (l  +  3a)2  =  (5-z)2  +  4(l-a)(3-2a;). 

15.  5  (2  r  +  7)  (r  -  2)  -  6  (r  +  4)2  =  5  +  (2  r  +  3)2. 

37.  Discussion  of  the  Axioms.  Two  equations  are  said  to  be 
Equivalent  when  the  roots  of  either  are  the  roots  of  the  other. 

As  remarked  in  paragraph  35,  when  the  number  used  as  an 
addend,  subtrahend,  multiplier,  or  divisor  is  an  arithmetical 
number,  then  the  given  equation  and  the  resulting  equation 
are  equivalent. 

Axioms  (a)  and  (b).  If  the  number  added  to  or  subtracted 
from  both  members  of  an  equation  is  an  expression  involving  the 
unknown  number,  the  given  equation  and  the  resulting  equation 
are  equivalent,  provided  the  expression  has  a  finite  value  for  the 
root  or  roots  of  the  equation. 

Axiom  (c).  If  the  multiplier  is  an  expression  involving  the 
unknown  number,  the  new  equation  may  not  be  equivalent  to  the 
given  equation. 

Example.     Sx  —  2  =  x  —  1  has  the  root  \. 

Multiplying  both  members  by  x  —  2, 

Sx2-8x  +  4=x2-Sx-\-2,  or2x2-5x-\-2  =  0. 

This  equation  has  the  root  2  f  or  2  .  22  -  5  •  2  +  2  =  8  -  10  +  2  =  0. 
The  given  equation  does  not  have  the  root  2,  for  3-2  —  2  does  not 
equal  2  —  1. 

Hence,  multiplying  the  given  equation  by  x  —  2  introduces  the  root  2. 

When  the  multiplier  is  an  arithmetical  number,  or  is  the 
lowest  common  multiple  of  the  denominators  of  a  fractional 
equation,  the  resulting  equation  and  the  given  equation  are 
equivalent. 


SIMPLE   EQUATIONS  37 

Axiom  (d).  If  the  divisor  is  an  expression  involving  the  un- 
known number,  the  new  equation  and  the  given  equation  are  not 
equivalent. 

Example,     x2  -  4  =  x  -  2  has  the  roct  2,  for  22  -  4  =  2  -  2. 

Dividing  both  members  by  x '•  —  2,  x  +  2  =  1. 

This  equation  does  not  have  the  root  2,  for  2  +  2  does  not  equal  1. 

Whenever  an  equation  is  divided  by  an  expression  involving 
the  unknown  number,  one  or  more  roots  of  the  given  equation 
are  lost.  The  following  example  illustrates  a  common  instance 
of  this  fact. 

Example.     3  x2  —  2  x  =  0. 

Bx  :  3  x  -  2  =  0,  or  x  =  }. 

|  is  indeed  a  root  of  the  equation  3  x2  —  2  x  =  0,  but  it  is  not  the  only 
root,  x  =  0  is  also  a  root,  for  3  •  02  —  2  •  0  =  0.  This  root  is  obtained  by 
setting  the  divisor  x  equal  to  zero. 

In  general,  if  the  expression  by  which  both  members  of  an 
equation  is  divided  is  set  equal  to  zero,  the  roots  of  the  result- 
ing equatipn  are  also  roots  of  the  given  equation. 

38.    Mechanical  Processes  of  Solving  Equations. 

(a)  Transposition.  A  term  may  be  transposed  from  one  mem- 
ber of  an  equation  to  the  other  provided  its  sign  is  changed. 

Proof.     Let  x  4-  a  =  b. 

S0  :  x  =  b  —  a.  Axiom  (?>),  §  35 

(b)  Cancellation.  A  term  which  appears  in  both  members  of 
an  equation  may  be  cancelled. 

Proof.     Let  x  +  a  —  b  +  a. 

Sa:  x  =  b.  Axiom  (6),  §35 

(c)  Changing  Signs  in  an  Equation.  The  signs  of  all  of  the 
terms  of  an  equation  may  be  changed. 

Proof.     Let  ax  —  b  =  c  —  dx. 

M_i :  —  ax  -f-  b  =  —  c  +  dx.  Axiom  (c),  §  35 


38  ALGEBRA 

(d)  Clearing  of  Fractions.     An  equation  may  be  cleared  of 

fractions  by  multiplying  both  members  by  the  lowest  common 

denominator  of  the  fractions  involved.     This  process  is  based 

upon  axiom  (c),  §  35. 

5                8 
Example.     Solve  the  equation =  0 

4a-3      7#-3 

Solution:  1.   M^x-sh7x-3)  :  5(7  x  —  3)  —  8(4  x  —  3)  =  0. 

2.  .-.  86«  —  16  —  32*  + 24  =  0. 

3.  .-.  8*  =  ~9,ory=-  3. 

Check  :  Does =  0  ?     Does  — —  =  0  ? 

_  12  -  3       -  21  -  3  -  15      -  24 

Does  (-t)_(_i)B0f     Yes. 

EXERCISE  12 
Solve  the  following  equations  : 

i  i_J_  =  i_JL  4   1.    JL    A.     X' 

2     9  a     9       6af  5<     10*      15 1         12' 

-      2  _?  +  A  =  i_ii  -     2a-7       10*-3 


3  y      y     2  y  6y  x2  —  4      5  #(#  +  2) 

2iaH-7a  +  ll  =  3  _27_         8  18 

7a2_4a_9  '  '   ^_5     z  +  2     Z2_sz-10' 

7       7m  5m    _   12(m2-l) 

m  +  3      m  —  1      m2  +  2  m  —  3 

8.    -^ 3U.+ 2 =  0. 

2r  +  l      3r  +  2     6r2+7r  +  2 

12*  — 5       3s+4     _4s-5 


10. 


•21  3(3«+l)  7 

3  w  -  5      4w  +  2==15w-l      7 

2  3w  +  2  10  5 


11        «*  +  3  1        =        2a?-l 

'    2(s»-8)      6(a>-2)      3(x2  +  2x  +  4) 


SIMPLE   EQUATIONS  39 

12.  .05  a; -1.82 -.7  a;  =  .008  a; -.504. 

13.  2.88  y  -  .756  +  .62  y  -  .858  =  .81  y. 

14     *-3      f  +  4=  St  +  20 
f.+  l     I  — 2     *2-£-2 

Solve  the  following  equations  for  cc: 

15-  4  +  r+-=a  +  6  +  c- 

«o     oc     ac 


3m 


2a;  +  5 

m     3a;  — 

■4m      6 a2  +  7  ma;  — 

20  m2 

3a; 

2x  +  n 

.T-f-2  7i 

+     2a 

=  2. 

x  +  a 
x-2a 

x—  a 
»  +  3  a 

2«a;-19a2  _Q 
ar*  +  ax  —  6  a2 

x(a  +  4=b)-b2  , 

x  —  b  __  x  +  a 

a2- 

-&2 

a-\-b      a—b 

a 

b 

a  —  b 

x  +  b 

x-\-  a     x 

+  a  +  b 

16. 
17. 
18. 
19. 
20. 


39.  Algebraic  Translation.  In  applications  of  algebra,  number 
relations  expressed  in  words  must  be  expressed  by  means  of 
algebraic  symbols.  This  process  may  be  termed  "  translation." 
Skill  in  making  such  translation  depends  in  part  upon  care  in 
reading  the  statement  which  gives  the  number  relations  and 
in  part  upon  familiarity  with  a  few  simple  devices.  For  the 
elementary  instruction  preparatory  to  the  solution  of  the  fol- 
lowing review  exercises,  see  the  "  First  Year  Algebra." 

Example.  Express  in  symbols :  the  sum  of  the  squares  of 
two  given  numbers  decreased  by  4  times  their  quotient. 

Solution  :  1.    Let  x  and  y  be  the  given  numbers. 

2.  Then  x2  and  y2  are  the  squares  of  these  numbers,  and  -  is  their 

quotient.  * 

x 

3.  The  expression  is  :  x2  +  y'2  —  4  -  • 


40  ALGEBRA 

EXERCISE  13 
Express  in  symbols  the  following : 

1.  Five  times  a  certain  number. 

2.  The  sum  of  the  cubes  of  two  given  numbers. 

3.  3  more  than  five  times  a  given  number. 

4.  The  excess  of  10  over  a  given  number ;  of  y  over  5. 

5.  5  less  than  three  times  a  given  number;  7b  diminished 
by  c. 

6.  The  amount  by  which  15  exceeds  twice  a  given  number. 

7.  The  difference  between  5  and  13 ;  between  a  and  20. 

8.  Five  per  cent  of  x  dollars ;  a  per  cent  of  D  dollars. 

9.  The  simple  interest  on  P  dollars  for  four  years  at  r  per 
cent. 

10.  The  amount  to  which  M  dollars  accumulates  in  t  years 
when  invested  at  five  per  cent  simple  interest. 

11.  The  number  by  which  3x  exceeds  (x—  6). 

12.  The  larger  part  of  18  if  s  is  the  smaller  part. 

13.  The  smaller  part  of  3  x  if  (x  +  5)  is  the  larger  part. 

14.  The  larger  of  two  numbers  if  c  is  the  smaller  and  if  the 
difference  between  them  is  15. 

15.  The  smaller  of  two  given  numbers  whose  difference  is  4 
if  the  larger  is  y. 

16.  The  smaller  part  of  35  if  the  larger  part  is  x. 

17.  The  two  integers  consecutive  to  the  integer  represented 
by  m. 

18.  The  two  odd  integers  consecutive  to  x:    (a)  if  x  is  an 
odd  integer ;    (6)  if  x  is  an  even  integer. 

19.  The  complement  of  a  degrees ;  the  supplement. 

20.  The  third  angle  of  a  triangle  of  which  the  other  two  are 
angles  of  x  degrees  and  (x  —  4)  degrees  respectively. 


SIMPLE   EQUATIONS  41 

21.  The  perimeter  of  a  rectangle  whose  base  exceeds  its  alti- 
tude by  3  inches. 

22.  The  ages  of  A  and  B  8  years  ago  if  A's  age  now  is  twice  B's. 

23.  The  difference  between  one  fifteenth  and  one  third  of  a 
certain  number. 

24.  The  total  number  of  cents  in  a  sum  of  money  consisting 
of  a  certain  number  of  dollars,  twice  as  many  quarters,  and 
three  times  as  many  dimes  as  quarters. 

25.  The  time  required  by  a  train  for  a  trip  of  A  miles : 
(a)  at  30  miles  an  hour;    (b)  at  r  miles  at  hour. 

26.  The  rate  at  which  an  automobile  travels  if  it  goes  D 
miles  :  (a)  in  7  hours ;  (b)  in  n  hours. 

27.  If  the  rate  of  a  river  is  3  miles  an  hour,  express  the 
time  required  by  a  boat  whose  rate  in  still  water  is  x  miles  an 
hour :  (a)  for  a  trip  of  20  miles  downstream  ;  (b)  for  a  trip  of 
20  miles  upstream  ;  (c)  for  a  trip  of  20  miles  downstream  and 
back  again. 

28.  The  area  of  a  parallelogram  whose  base  is  3  feet  less 
than  twice  its  altitude. 

29.  The  part  of  a  piece  of  work  a  man  can  do :  (a)  in  c  days, 
if  he  can  do  all  of  it  in  10  days ;  (b)  in  3  days,  if  he  can  do  all 
of  it  in  x  days. 

30.  The  reciprocal  of  3  x. 

31.  The  number  whose  hundreds'  digit  is  a,  whose  tens' 
digit  is  b,  and  whose  units'  digit  is  c. 

32.  The  number  whose  digits  are  the  same  as  those  of  the 
number  in  Example  31,  but  in  the  reverse  order. 

Express  in  words  the  following  expressions : 


33.    \ab. 

35.    x2  +  y2  —  2  xy. 

37.    a2-{-b2. 

34.    2(x  +  y)  +  3(x-y). 

36.  }m(w-f»). 

38.    (a3  +  63)3. 

39.    *±*. 

X2y2 

40.    fO+32. 

42 


ALGEBRA 


40.  A  Problem  is  a  statement  of  one  or  more  relations  be- 
tween one  or  more  unknown  numbers  and  certain  known  num- 
bers, from  which  the  unknown  numbers  are  to  be  determined. 

In  this  paragraph  certain  problems  are  considered  which  can 
be  solved  by  using  only  one  unknown  number. 

Example.  The  rate  of  an  express  train  is  five  thirds  of  that 
of  a  slow  train.  It  travels  75  miles  in  one  hour  less  time  than 
the  slow  train.     Find  the  rate  of  each  train. 

Solution  :  1.  Let  r  =  the  no.  of  mi.  in  the  rate  per  hour  of  the  slow 
train. 

2.   Then  |  r  =  the  no.  of  mi.  in  the  rate  per  hour  of  the  fast  train. 


8. 


Hence 

for  one  train 

for  the  other 

the  rate  is 
r 

the  distance  is 
75 
75 

the  time  is 

75 

r 

75  +  fr 

4.   .-.  ^  = 


1. 


75 
fr       r 
When  this  equation  is  solved,  r  is  found  to  equal  30. 
Hence  the  rate  of  the  slow  train  is  30  mi.  an  hour,  and  of  the  fast  train, 
therefore,  50  mi.  an  hour. 

Check  :  The  time  for  the  slow  train  for  75  mi.  is  75  +  30  or  2.5  hr. 
The  time  for  the  fast  train  for  75  mi.  is  75  -f-  50  or  1.5  hr. 
The  time  of  the  latter  is  one  hour  less  than  that  of  the  former. 


EXERCISE  14 

1.   The  denominator  of  a  certain  fraction  exceeds  its  numer- 


ator by  6.  If  the  numerator  be  increased  by  7  and  the  denomi- 
nator be  decreased  by  5,  the  fraction  becomes  i73-.  Find  the 
fraction. 

2.  Divide  55  into  two  parts  whose  quotient  shall  be  4. 

3.  Divide  54  into  two  parts  such  that  twice  the  smaller 
shall  exceed  29  as  much  as  143  exceeds  four  times  the  greater. 


SIMPLE   EQUATIONS  43 

4.  The  perimeter  of  a  certain  rectangle  is  330  feet.  The 
altitude  is  four  sevenths  of  the  base.     Find  the  dimensions. 

5.  The  numerator  of  a  certain  fraction  exceeds  the  de- 
nominator by  5.  If  the  numerator  be  decreased  by  9,  and  the 
denominator  be  increased  by  6,  the  sum  of  the  resulting  frac- 
tion and  the  given  fraction  is  2.     Find  the  fraction. 

6.  Divide  197  into  two  parts  such  that,  when  the  greater 
is  divided  by  the  smaller,  the  quotient  is  5  and  the  remainder 
is  23. 

7.  The  length  of  a  certain  lot  is  three  times  its  width.  If 
the  length  be  decreased  by  20  feet  and  the  width  be  increased 
by  10  feet,  the  area  will  be  increased  by  200  square  feet.  Find 
its  present  dimensions. 

8.  If  one  fifth  of  the  supplement  of  a  certain  angle  be 
diminished  by  two  elevenths  of  the  complement  of  the  angle, 
the  result  is  19.     Find  the  angle. 

9.  A  passenger  train  whose  rate  is  35  miles  an  hour  and  a 
freight  train  whose  rate  is  25  miles  an  hour  start  at  the  same 
time  from  points  which  are  100  miles  apart,  (a)  If  they  travel 
toward  each  other,  in  how  many  hours  will  they  meet  ?  (6)  If 
they  travel  away  from  each  other,  in  how  many  hours  will  they 
be  150  miles  apart  ? 

10.  A  freight  train  runs  6  miles  an  hour  less  than  a  pas- 
senger train.  It  runs  80  miles  in  the  same  time  that  the 
passenger  train  runs  112  miles.     Find  the  rate  of  each. 

11.  A  man  walks  10  miles  and  then  returns  in  a  carriage 
whose  rate  is  3  times  as  great  as  his  rate  of  walking.  If  it 
took  him  4  hours  less  time  returning  than  going,  what  was  his 
rate  of  walking  ? 

12.  A  vessel  runs  at  the  rate  of  12  miles  an  hour.  If  it 
takes  as  long  to  run  27  miles  upstream  as  45  miles  downstream, 
what  is  the  rate  of  the  current  ? 


44  ALGEBRA 

13.  A  man  travels  130  miles  in  ten  hours  in  an  automobile, 
part  of  the  distance  at  an  average  rate  of  20  miles  an  hour  and 
the  rest  at  an  average  rate  of  10  miles  an  hour.  How  far  does 
he  travel  at  each  rate  ? 

14.  A  man  receives  $  140  per  year  as  interest  on  $  2500. 
$  500  is  invested  at  5  %  ;  part  of  the  remainder  at  5i  %  ;  and 
the  rest  at  6  %.     How  much  has  he  invested  at  5^  %  ? 

15.  A  man  travels  24  miles  at  the  rate  of  5  miles  an  hour. 
By  how  many  miles  an  hour  must  he  increase  his  rate  in  order 
to  make  the  trip  in  one  fourth  of  the  time  ? 

16.  Find  three  consecutive  numbers  such  that  the  square  of 
the  greatest  shall  exceed  the  product  of  the  other  two  by  49. 

17.  Find  the  upper  base  of  a  trapezoid  whose  area  is  175 
square  inches,  whose  altitude  is  10  inches,  and  whose  lower 
base  is  20  inches. 

18.  Two  automobilists  use  gasoline  from  a  tank  containing 
60  gallons.  If  one  uses  gasoline  at  the  rate  of  live  gallons  in 
three  days  and  the  other  five  gallons  in  seven  days,  how  long 
will  the  60  gallons  last  ? 

19.  If  A  can  do  a  piece  of  work  in  5  days,  B  in  8  days,  and 
C  in  10  days,  how  many  days  will  it  take  them  to  do  the  work 
if  they  work  together  ? 

20.  A  man  has  $  6.90  in  dollars,  half  dollars,  and  dimes. 
The  number  of  halves  is  twice  the  number  of  dollars,  and  the 
number  of  dimes  is  equal  to  the  sum  of  the  number  of  dollars 
and  the  number  of  half  dollars.  Find  the  number  of  coins  of 
each  kind. 

SUPPLEMENTARY  PROBLEMS 

21.  Divide  a  into  two  parts  whose  quotient  shall  be  m. 

22.  The  numerator  of  a  certain  fraction  exceeds  its  denomi- 
nator by  m.  If  the  denominator  be  increased  by  n,  the  fraction 
becomes  f.  (a)  Find  the  fraction,  (b)  Find  the  value  of  the 
fraction  when  m  is  4  and  n  is  2. 


SIMPLE   EQUATIONS  45 

23.  The  perimeter  of  a  certain  rectangle  is  c  feet.  The  base 
exceeds  the  altitude  by  d  feet,  (a)  Find  the  dimensions  of  the 
rectangle.  (6)  Find  the  values  of  the  dimensions  when  c  equals 
50,  and  d  equals  5. 

24.  The  length  of  a  certain  field  is  m  times  its  width.  If 
the  length  be  increased  by  r  feet,  and  the  width  by  s  feet,  the 
area  will  be  increased  by  t  square  feet,  (a)  Find  the  dimen- 
sions of  the  field.  (b)  Find  the  values  of  the  dimensions  when 
m  is  4,  and  r,  s,  and  t  are  respectively  2,  3,  and  48. 

25.  Divide  c  into  two  parts  such  that  the  sum  of  one  mth 
of  the  first  part  and  one  nth  of  the  second  part  shall  equal  d. 
Find  the  values  of  the  parts  when  m  is  2  and  n  is  3. 

26.  If  A  can  do  a  piece  of  work  in  a  days,  B  in  b  days,  C  in 
c  days,  and  D  in  d  days,  how  many  days  will  it  take  them  to 
do  the  work  if  all  work  together  ? 

27.  A  sum  of  money  amounting  to  m  dollars  consists  entirely 
of  quarters  and  dimes,  there  being  n  more  dimes  than  quarters. 
How  many  are  there  of  each  ? 

28.  At  what  time  between  8  and  9  o'clock  will  the  hands  of 
a  clock  be  together  ? 

29.  At  what  time  between  2  and  3  o'clock  is  the  minute 
hand  of  a  watch  15  minute  spaces  ahead  of  the  hour  hand  ? 

30.  In  a  mixture  of  sand  and  cement  containing  one  cubic 
yard,  16  %  is  cement.  How  much  sand  must  be  added  to  the 
mixture  so  that  the  resulting  mixture  will  contain  12  °J0  of 
cement  ? 


V.     GRAPHICAL   REPRESENTATION 

41.  In  the  figure  below:  XX'  is  called  the  Horizontal  Axis; 
YY'  is  called  the  Vertical  Axis ;  together  they  are  called  the 
Axes ;  the  point  0  is  called  the  Origin ;  PR,  perpendicular  to 
the  horizontal  axis,  is  called  the  Ordinate  of  the  point  P;  PS, 
perpendicular  to  the  vertical  axis,  is  called  the  Abscissa  of  P; 
PR  and  PS  together  are  called  the  Coordinates  of  P.  Distances 
on  OX  are  considered  positive,  on  OX'  negative,  on  OF  posi- 
tive, and  on  OY'  negative.  The  part  of  the  plane  within  the 
angle  XOY  is  called  the  first  quadrant;  the  part  within  the 
angle  YOX'  is  called  the  second  quadrant;  etc.  The  abscissa 
of  P,  according  to  the  indicated  scale,  is  3,  and  the  ordinate 
is  4.     The  point  P  is  called  the  Point  (3,  4). 


Y 

_ 

1 

1  i" 

J., 

rp 

!, 

! 

„ 

! 

i 

, 

L 

xn 

1 

O 

Ik 

X 

!  !-fi 

-4 

- 

1 

♦ 

+ 

] 

♦4 

»4 

♦S 

+ 

~t 

1 

, 

1     \ 

1 

1 

&• 

, 

1 

1 

j* 

1 

- 

,  1 

' 

V 

46 


GRAPHICAL    REPRESENTATION 


47 


EXERCISE   15 

1.  What  are  the  coordinates  of  each  of  the  points  in  the 
figure  above  ? 

2.  Locate  (plot)  on  a  similar  diagram  the  following  points : 
(a)  (0,5);  (6)  (-3,4);  (c)  (-5,0);  (<i)  (0,  -4);  (e)  (-4,  -6). 

3.  In  which  quadrant  does  a  point  lie :  (a)  whose  abscissa 
is  positive  and  whose  ordinate  is  negative?  (b)  whose  abscissa 
and  whose  ordinate  are  both  negative  ? 

4.  What  sign  does  the  abscissa  of  a  point  have  if  the  point 
is  in  the  fourth  quadrant  ?  in  the  second  quadrant  ? 

5.  Change  the  word  "  abscissa  "  in  Example  4  to  "  ordinate," 
and  answer  the  resulting  questions. 

6.  Locate  the  points  whose  coordinates  are  given  in  the 
following  table ;  connect  the  points  by  a  smooth  curve  and  thus 
obtain  a  graph  showing  the  relation  between  the  ordinate  and 
the  abscissa  of  a  point  on  the  graph. 


X 

y 

-5-4 

-3 

-2 

-  1 

0 

1 

2. 

3 

4 

5 

25         16 

9 

4 

1 

0 

1 

4 

0 

16 

25 

Express  the  relation  between  y  and  x  by  means  of  an  equa- 
tion. 

7.  Water,  falling  from  any  height,  exerts  a  pressure  depend- 
ing upon  the  height  from  which  the  water  falls.  This  height 
is  called  the  "head"  of  the  water.  Draw  a  graph  showing 
the  relation  between  the  head,  expressed  in  feet,  and  the  pres- 
sure, expressed  in  pounds  per  square  inch,  using  the  data  given 
in  the  following  table. 

(Hint  :  use  the  head  for  the  abscissa  and  the  pressure  for  the  ordinate 
of  the  point. ) 


Head 

1 

2 

5 

8 

10 

20 

Pressure 

.4 

.8 

2.1 

3.4 

4.3 

8.6 

48  ALGEBRA 

8.  From  the  graph  in  Example  7,  determine  the  pressure 
from  a  head  of :  (a)  6  feet ;  (b)  15  feet. 

9.  From  the  graph  in  Example  7,  determine  the  head  neces- 
sary to  give  a  pressure  of:  (a)  6  pounds  per  square  inch; 
(6)  of  7  pounds  per  square  inch. 

42.  Equations  Having  Two  Unknowns.  A  Solution  of  an 
equation  having  two  (or  more)  unknowns  is  a  set  of  values  of 
the  unknowns  which  satisfies  the  equation. 

Thus,  x  =  1  and  y  =  4  is  a  solution  of  x  •+•  y  =  5,  for  1+4=5;  also, 
x  =—  8  and  y  =  13  is  a  solution,  for  (—  8)  +  13  =  5. 

43.  For  every  value  of  one  unknown,  a  value  of  the  other 
may  be  determined. 

In  x  +  y  =  5  :  when  x  —  —  f,  (—  f)  +  y  =  5.     Hence  y  =  5  +  f  =  6f. 

An  equation  having  two  (or  more)  unknowns  has  an  infinite 
number  of  solutions ;  for  this  reason,  such  equations  are  called 
Indeterminate  Equations.  As  x,  in  such  an  equation,  changes 
in  value,  y  also  changes  in  value,  x  and  y  are  called  Variables 
and  the  equation  is  called  an  equation  having  two  variables. 

EXERCISE  16 

1.  Determine  by  substitution  which  of  the  following  pairs 
of  numbers  are  solutions  of  the  equation  2  x  —  3  y  =  10 : 

(a)  x  =  2,y  =  -2;  (b)  x  =  3,y  =  l;  (c)  x  =  \,y  =  -Z. 

2.  Determine,  as  in  §  43,  three  more  solutions  of  the  equa- 
tion given  in  Example  1. 

3.  Determine  six  solutions  of  the  equation  y  =  #2  —  10,  three 
for  positive  values  of  x,  and  three  for  negative  values  of  x. 
Plot  the  points  whose  coordinates  correspond  to  these  solutions 
and  connect  them  by  a  smooth  curve,  —  thus  obtaining  a  part 
of  the  graph  of  the  equation. 


GRAPHICAL   REPRESENTATION  49 

4.  F=ma  is  an  equation  encountered  in  physics.  Assume 
that  m  has  the  constant  value  50 ;  determine  ten  solutions  of 
the  resulting  equation,  locate  the  points  corresponding,  and 
draw  the  graph. 

(Hint  :  use  2^  as  ordinate  and  a  as  abscissa.) 

44.  If  a  rational  and  integral  monomial  (§  11)  involves  two 
or  more  letters,  its  degree  with  respect  to  them  is  denoted  by  the 
sum  of  their  exponents. 

Thus,  2  a2bxy3  is  of  the  fourth  degree  with  respect  to  x  and  y. 

45.  If  each  term  of  an  equation  containing  one  or  more  un- 
known numbers  is  rational  and  integral,  the  Degree  of  the 
Equation  is  the  degree  of  its  term  of  highest  degree. 

Thus,  if  x  and  y  denote  unknown  numbers  : 

ax  —  by  =  6  is  of  the  first  degree ; 
2  x2  -  3  xy2  =  5  is  of  the  third  degree. 

46.  In  a  later  course  in  mathematics,  the  graph  of  an  equa- 
tion of  the  first  degree  having  two  unknowns  is  proved  to  be  a 
straight  line.     This  line  may  be  found  by  the 

Rule.  —  1.    Determine  two  solutions  (§  42)  of  the  equation. 
2.   Plot  the  points  whose  coordinates  correspond  to  these  solu- 
tions and  draw  the  straight  line  determined  by  them. 

Note  1.     Do  not  have  the  two  points  too  close  together. 

Note  2.  The  coordinates  of  every  point  on  the  graph  satisfy  the 
equation  of  the  graph;  and  every  point  whose  coordinates  satisfy  the 
equation  must  lie  on  the  graph.  In  geometrical  language,  the  graph  is 
the  locus  of  points  whose  coordinates  satisfy  the  equation. 

47.  An  equation  of  the  first  degree  is  called  a  Linear 
Equation. 

Example.     Draw  the  graph  of  3  x  —  4  y  =  12. 

Solution:  1.    When  x  =  0,    3  .  0  —  4  y  =  12,    —  4y=12,  ov  y  =—  3. 
When  £=-4,  8(-4)-4|s  12,  -  12  -  4  y  =  12,  -  4  y  =  24,  or  y  =  —Q. 
2.    In  the  figure  below  the  points  are  located  and  the  line  is  drawn. 


50 


ALGEBRA 


Y 

"^                                                         > 

X                                                                         6~                                y^              -       X 

-6     -S    -4     -3   r9     "              +      +  Q    +3>4    +  5    +0 

.                         >^ 

i^      ^ 

2         ^ 

+     ^ 

7 

^^i     i 

^    -^ 

;2      jp 

-     E-*i-»r-     -          H-JS 

z:               2 

EXERCISE  17 
Draw  the  graphs  of  the  following  equations : 

1.  2x+3y  =  12.  3.    3  x  +  2y=  0. 

2.  3  aj  -  5  i/  =  30.  4.   4  a  -f  5  y  =  24. 

5.    (a)  Draw  the  graph  ofsc+2y  =  — 1. 

(b)  Multiply  both  members  of  the  original  equation  by  3 
and  draw  the  graph  of  the  resulting  equation. 

(c)  Multiply  both  members  of  the  original  equation  by  —  5 
and  again  draw  the  graph  of  the  equation  resulting. 

(d)  From  the  graphs  obtained  in  parts  (a),  (6),  and  (c),  what 
do  you  conclude  is  the  effect  upon  the  graph  of  an  equation 
of  multiplying  both  members  of  the  equation  by  the  same 
number  ? 

(e)  What  effect  does  it  have  upon  the  solutions  of  the 
equation  ? 


VI.     SIMULTANEOUS   LINEAR   EQUATIONS 

48.  Two  equations,  each  containing  two  (or  more)  unknowns, 
are  said  to  be  Independent  Equations  if  each  has  solutions  which 
are  not  solutions  of  the  other. 

49.  Two  independent  equations  which  have  one  (or  more) 
common  solutions  are  called  Simultaneous  Equations. 

50.  Two  independent  linear  equations  which  do  not  have  a 
common  solution  are  called  Inconsistent  Equations. 

51.  Graphical  Solution  of  Simultaneous  Linear  Equations  •Having 
Two  Unknowns. 

Rule.  —  1.   Draw  upon  one  sheet  the  graphs  of  both  equations. 

2.  Find  the  coordinates  of  the  point  common  to  the  two  lines. 
These  coordinates  give  the  common  solution.  The  solution  may 
be  checked  by  substitution. 

(  5  x  —  3  v  =  19.  (1) 

Example.     Solve  the  equations   J 

l7«+4jr-**  (2) 

Solution  :  1.    For  equation  (1)  :  if  x  =  5,  y  =  2  ;  if  x  =  —  1,  y  =  —  8. 

2.  For  equation  (2)  :  if  x  =  -  2,  y  =  4  ;  if  x  =  4,  y  =  —  6|. 

3.  The  graph  follows.  • 

4.  The  straight  lines  intersect  in  the  point  (2,  —  3). 
The  common  solution  is  x  —  2,  y  =  —  3. 

Check  :  In  (1 )  :  does  5  •  2  -  3(  -  3)  =  19  ?     Yes. 
In  (2)  :  does  7  •  2  +  4(-  3)  =  2  ?     Yes. 

Note  1.  The  solution  obtained  by  this  method  is  usually  only  an  ap- 
proximate solution  owing  to  the  impossibility  of  determining  exactly  the 
coordinates  of  the  point  of  intersection  of  the  lines. 

Note  2.  If  the  equations  are  inconsistent,  the  lines  will  be  parallel ; 
if  the  equations  are  dependent,  the  lines  will  coincide. 

51 


52 


ALGEBRA 


y 

,„ 

5     tK)          :  :~i 

\                    S 

V                  t 

X                    7 

\^l            -/ 

\I              / 

\                   4 

5                  / 

^o      2 

X    "                              r^  cr       r                           X 

-in          -5              \      /  +  5           +A) 

L_    > 

Sc«: it 

A-t 

*,/ V 

j!Z     I               it 

v       T 

J-         \ 

tE       \ 

-An                -S 

23^                   £                   it 

^ 

t         II 

Y 

EXERCISE  18 

Study  the  following  pairs  of  equations  graphically  ;  if  simul- 
taneous, determine  their  common  solutions  : 


1. 


3x  +  y  =  11. 

5  x  —  y  =  13. 


2  x  +  5y  =  10. 
4«  +  10y  =  40. 


2. 


4^  +  3  ?/  =  -  1. 

5x  +  2/  =  7. 


6. 


3a;-2y  =  6. 
9s-6y  =  18. 


3. 


7  x  +  8y  =  26. 


7.   ; 


5  a;  —  4  ?/  =  0. 

7  as  +  G?/  =  —  29. 


5  a?  +  3  y  =  14. 

4  #  —  5  ?/  =  26. 


8. 


9  x  +  14  ?/  =  -  25. 
3  a,  _  4  y  =  22. 


52.  Two  simultaneous  equations  having  two  variables  may 
be  solved  by  combining  them  so  as  to  cause  one  of  the  variables 
to  disappear.     This  process  is  called  Elimination. 


SIMULTANEOUS   LINEAR   EQUATIONS  53 

53.    Elimination  by  Addition  or  Subtraction. 

Rule.  —  1.  Multiply,  if  necessary,  both  equations  by  such  numbers 
as  will  make  the  coefficients  of  one  of  the  variables  of  equal  absolute 
value. 

2.  If  the  coefficients  have  the  same  sign,  subtract  one  equation 
from  the  other ;  if  they  have  opposite  signs,  add  the  equations. 

3.  Solve  the  equation  resulting  from  step  2  for  the  other  variable. 

4.  Substitute  the  value  of  the  variable  found  in  step  3  in  any 
equation  containing  both  variables,  and  solve  for  the  remaining 
variable. 

5.  Check  the  solution  by  substituting  it  in  both  of  the  original 
equations. 


Example.     Solve  the  equal 

.         (5a>  +  3y  =  -9. 
;ions  \  n          .              ._ 
[  3  x  -  4  y  =  -  17. 

(1) 
(2) 

Solution:    1.    ilf4*(l): 

20  x+  12?/ =-30. 

(3) 

2.  J6(S)i 

9x-12?/=-51. 

(4) 

3.    (3)  +  (4)  : 

29z=-87. 

(5) 

4. 

.-.  x=-S. 

5.    Substitute  —  3  for  x  in  (1) : 

-15  +  Sy  =-9. 

6. 

.-.  3  y  =  0,  or  y  =  2. 

The  solution  is :  x  =  —  3,  y  =  2. 

Check:   In  (2) :  does3(-3)- 

42  =-17?     Yes. 

In  (1):  does  5(-  3)  + 

3-2=-    9?     Yes. 

54.    Elimination  by  Substitution. 

Rule.  —  1.  Solve  one  equation  for  one  variable  in  terms  of  the 
other  variable. 

2.  Substitute  for  this  variable  in  the  other  equation  the  value 
found  for  it  in  step  1. 

3.  Solve  the  equation  resulting  in  step  2  for  the  second  variable. 

4.  Substitute  the  value  of  the  second  variable,  obtained  in  step  3, 
in  any  equation  containing  both  variables  and  solve  for  the  first 
variable.  ' 

5.  Check  the  solution  by  substituting  it  in  the  original  equations. 

*  M4  (1)  :  means  "  Multiply  both  members  of  equation  (1)  by  4." 


54  ALGEBRA 


Example.     Solve  the  equations 


11  x-  5  y  =  4.  (1) 

4  x  -  3  y  =  5.  (2) 


Solution:    1.    Solve  (1)  for  y :  y  —  llx — !,        (3) 

5 

2.  Substitute  in  (2)  :  4x  -  3  [  1!  x  ~  4>j  %  5.  (4) 

3.  Solving  (4)  f or  x  :  20  x  -  33  x  +  12  =  25  ; 

-  13  x  =  13,  or  se=-  1. 

4.  Substitute  —  1  for  x  in  (1)  :  —  11  —  5  y  =  4. 

.  •.  —  5  ?/  =  15,  or  y  =  —  3. 
The  solution  is  :  x  =—  1,  y  =  —  3. 

Check  :  In  (1)  :  does  11(-  1)  -  5(-  3)  =  4  ?  does  -  11  +  15  =  4  ?  Yes. 
In  (2)  :     does4(- 1) -3(-3)  =5?    does  -  4  +  9    =  5  ?  Yes. 

EXERCISE   19 

Solve  the  following  pairs  of  equations  by  addition  or  sub- 
traction. (If  difficulty  is  experienced  in  obtaining  a  solution, 
determine  graphically  whether  the  equations  are  inconsistent 
or  dependent.) 

[  2  x  -  3  y  =  19.  (  6  x  +  11  y  =  31. 

1.  J  5.       * 

7#  +  4  y  =  23.  j  6  y  -  11  a;  =  74. 


2. 


a;  -  5  ?/  =  -  21.  (  3  x  +  2  w  =  -  31. 

6. 
\  3  a: -87  =-35.  1  6  a;  +  4y  =  -  62. 


f  15x  +  8y  =3.  [4  y_8w  =  -3. 

3.    •!  ^  ,-^  7. 


6z-12?/  =  5.  Ill  2*4-5  w== -15. 

f  13  m  -  7  w  =  15.  [  3  x  -  4  y  =  -  13. 

4.  .8. 

[  8  m  -  4  n  =  9.  }  6  a,-  -  8  ?/  =  -  5. 

Solve  the  following  pairs  of  equations  by  substitution: 

2x+y  =  8.  Ja  +  26  =  11. 

Waj-4y  =  43.  10"    {  3a -f- 5  6  =  29. 


SIMULTANEOUS  LINEAR   EQUATIONS 


55 


11. 


12. 


3r4-7s  =  -12. 
_6r  +  9s  =  l. 

8e-3/=47. 
6e-7/=21. 


13. 


14. 


5x  -{-  6y  =  —  5. 
10  a  +  9  y  =  -  6. 

3  x  -  5  y  =  38. 
-5x4-3?/  =  - 26. 


Solve  the  following  pairs  of  equations  in  either  manner : 


15. 


16. 


17. 


18. 


19. 


6^~  9 


6   '        2 

9 


lla-36  =  3a  +  fe 
11  8 

8  a  -5  6  =  1. 


2  e  +  *  +  6 
e-2t-3 

5  e 


3t— 7. 

I 

4r— 3s     r-6g 

14  9 

2r  +  3«  =  -10. 


8 


a?—  3     y  —  5 
9  5 

2aj-l     3#  +  4 

d-2n 


20. 


I3  <*  4-  n  4.  3 

(  d  4-  4  »  —  7 


=  0. 

1 
5' 


11 


21. 


»     y 

8_9  =  7 
*      2/ 


Hint:    Eliminate  y  without  clear- 
ing of  fractions. 


22. 


23. 


\9-+ 

14_ 

11 

2  ' 

d 

21  _ 

s 

-7. 

8_ 

X 

3^a 

5y 

89 
30 

_5_ 
6x 

_6_ 
2/ 

59 
18 

'  2 
3a; 

3 
4/ 

1 
12* 

5 

4  a; 

4 

3/ 

13 

72' 

24. 


Solve   the    following    for   x 

and  ?/ : 

[  o  x  -  6  y  =  8  a. 
25.     i 

I  4  a;  4-  9 1/  =  7  a. 


56 


ALGEBRA 


26. 


27. 


28. 


bx  —  ay  =  b2. 

(a  —  b)x  +  by  =  a2. 

ax  +  by  =  1. 
cx+  dy  =  1. 

ax  +  by  =  2  a. 
a2x  —  b2y  =  a2-\-  b2. 


j  2a#  —  &y 


29. 


30. 


a 

x-\-by 
3a  +  2 

=  b. 

m 

n 

n  +  y 

m  — 

X 

m 

n 

n+x      m—y 


55.    Equations  Containing  Three  or  More  Variables. 
Example.     Solve  the  set  of  equations : 

12  m  —  4  n  +  p  =  3. 
m  —  n  —  2  2~>  =  -  1. 

5m-2n  =  0. 


Solution:   1.    M2  (1)  : 

2.  (2)  +  (4): 

3.  Mg  (3)  : 

4.  (5)  -(6): 

5.  Substitute  5  for  n  in  3. 


24w»-  8n  +  2j>  =  6. 

25  m  —  9  w  =  5. 

25  m  - 10  n  =  0. 

n  =  5. 
5  m  -  10  =  0,  or  to  =  2. 


(1) 
(2) 
(•3) 

(4) 

*(5) 
(6) 


6.    Substitute  5  for  n  and  2  for  to  in  (2)  : 

2_5-2^=-l. 

.-.  —  2p  =  2,  orp  =—  1. 
Solution  :  M».=  2,  n  =  5,  jj  =s—  1. 

Check  :  The  solution  satisfies  each  of  the  three  given  equations. 

Rule.  —  To  solve  a  system  of  three  equations  containing  three 
unknowns : 

1.  From  two  equations,  say  the  ist  and  2d,  eliminate  one  of 
the  unknowns  ;  mark  the  resulting  equation  as  equation  (4). 

2  From  another  pair  of  equations,  say  the  ist  and  3d,  eliminate 
the  same  unknown,  and  mark  the  resulting  equation  as  equa- 
tion (5). 

3.  Equations  (4)  and  (5),  containing  two  unknowns,  are  readily 
solved  (if  the  system  has  a  solution;.  Then  by  substitution  the 
third  unknown  may  be  determined. 


SIMULTANEOUS   LINEAR   EQUATIONS 


57 


EXERCISE   20 

Solve  the  following  sets  of  equations : 


3  x  +  y  —  z  =  14. 
x  +  3  y  —  z  =  16. 

_x  +  y  —  3z  =  —  10. 

3p  +  4(/  -f  5r  =  10. 

4  P  —  5  c/  —  3  r  =  25. 

5  p  —  3  </  —  4  r  =  21. 


4.x -11  y-  5  z  =  9. 
3.     {  8  x  -f-  4  ?/  —  z  =  11. 
16a?-f-72/-f6z  =  64. 

2  x-  +  4  ?/  —  z  =  —  2. 
j  i8a-8#  +  4z=  -25. 
10a,--}-4?/-9z  =  -30. 


5 

X 

8_ 

3. 

8 

-:5=i 

V 

?/ 

2 

2." 

2 

>       7 
+  3~x~~ 

=  2. 

3 

v.  -\-x  = 

-5. 

4 

x-y  = 

21. 

5 

y  +  z  = 

-19. 

6 

z  —  u  = 

39. 

Solve  Examples  6,  7,  and  8 
for  #,  y,  and  2. 


6. 


7. 


10. 


ax  -{-by  = 

a3+  &3 
abc 

by  +  cz  =  - 

9  +  c3 
abc 

cz  +  aa?  = 

c3  +  az 
abc 

[1      1 
-  +  -  =  a. 

#    y 

i    1    , 

y     z 

1      1 

-+-  =  c. 

Z        X 

b      a 

-  +  -  =  c. 
x      y 

c      6 

_  +  _  =  <*. 

.2/      » 

u  —x  -\-y 

=  15. 

x  —  y  +  z 

=  -12. 

y  —  z  +  u 

=  13. 

.z  —  u  +  X 

=  -14 

58  ALGEBRA 

56.  Solution  by  Formula.  Simultaneous  linear  equations  hav- 
ing two  or  more  unknowns  may  be  solved  by  means  of  certain 
formulae.  This  method  of  solution  is  considered  in  §  233,  page 
240,  and  may  be  studied  at  this  time  if  desired. 

57.  Many  problems  are  solved  more  conveniently  by  using 
two  or  more  unknowns. 

Example  1.  A  certain  number  of  two  digits  exceeds  three 
times  the  sum  of  its  digits  by  4.  If  the  digits  be  reversed,  the 
sum  of  the  resulting  number  and  the  given  number  exceeds 
three  times  the  given  number  by  2.     Find  the  number. 

Solution  :     1.   Let  t  =  the  tens'  digit,  and  u  =  the  units'  digit. 

2.  .-.  10  £  +  u  =  the  original  number, 

and  10  u  +  t  =  the  number  obtained  by  reversing  the 

digits. 

3.  .  \  10 1  +  u  =  3  (t  +  u)  +  4 

or  7t-2u  =  4.  (1) 

4.  Also  (10«+  m)+  (10m +0  =3  (101+  u)  +  2, 

or  19t-8tt  =  -2.  (2) 

5.  Solving  equations  (1)  and  (2)  by  the  usual  methods, 

t  =  2  and  u  =  5. 
.-.  25  is  the  number. 
Check:     Does  25  =  3  (2  +  5)  +  4  ?     Yes. 

Does  25  +  52  =  3  (25)  +  2  ?     Yes. 

EXERCISE  21 

1.  Divide  79  into  two  parts  such  that  twice  the  smaller  ex- 
ceeds the  greater  by  5. 

2.  If  3  be  added  to  the  numerator  of  a  certain  fraction,  and 
7  be  subtracted  from  the  denominator,  the  value  of  the  frac- 
tion becomes  -^-.  If  1  be  subtracted  from  the  numerator,  and  7 
be  added  to  the  denominator,  the  value  becomes  f.  Find  the 
fraction. 

3.  The  units'  digit  of  a  certain  number  of  two  digits  is  one 
third  of  the  tens'  digit.     If  the  number  is  divided  by  the  differ- 


QUADRATIC    EQUATIONS  59 

ence  of  its  digits,  the  quotient  is  15  and  the  remainder  is  3. 
Find  the  number. 

4.  Find  the  three  angles  of  an  isosceles  triangle  if  each  of 
the  base  angles  exceeds  the  vertical  angle  by  30°. 

5.  There  are  two  numbers  such  that  when  the  first  is  divided 
by  the  second  the  quotient  is  3  and  the  remainder  is  1 ;  when 
the  second  is  divided  by  one  fifth  of  the  first,  the  quotient  is  1 
and  the  remainder  is  3.     Find  the  numbers. 

6.  A  man  has  $10,000  invested,  part  at  5%  and  part  at  6%. 
The  interest  for  one  year  on  the  5%  investment  exceeds  the 
interest  for  one  year  on  the  6  %  investment  by  $  60.  How 
much  does  he  have  invested  at  each  rate  ? 

7.  A's  age  is  six  fifths  of  B's.  Fifteen  years  ago  his  age 
was  thirteen  tenths  of  B's.     Find  their  present  ages. 

8.  If  the  numerator  of  a  certain  fraction  be  increased  by 
4,  the  value  of  the  fraction  becomes  4 ;  if  the  denominator  of 
the  fraction  is  decreased  by  3,  the  value  of  the  fraction  be- 
comes -f.     Find  the  fraction. 

9.  A  certain  chord  of  a  circle  divides  the  circumference 
into  two  arcs  such  that  three  times  the  minor  arc  exceeds 
twice  the  major  arc  by  80°.     Find  the  two  arcs. 

10.  The  units'  digit  of  a  certain  number  of  two  figures  exceeds 
the  tens'  digit  by  5.  If  the  number,  increased  by  6,  be  divided 
by  the  sum  of  its  digits,  the  quotient  is  4.     Find  the  number. 

11.  A's  age  is  twice  the  sum  of  the  ages  of  B  and  C.  Two 
years  ago,  A  was  4  times  as  old  as  B,  and,  four  years  ago,  A 
was  6  times  as  old  as  C.     Find  their  ages. 

12.  Find  three  numbers  such  that  the  sum  of  the  first,  one 
half  of  the  second,  and  one  third  of  the  third  shall  equal  29  ; 
also  such  that  the  sum  of  the  second,  one  third  of  the  first, 
and  one  fourth  of  the  third  shall  equal  28;  and  finally  such 
that  the  sum  of  the  third,  one  half  of  the  first,  and  one  third 
of  the  second  shall  equal  36. 


60  ALGEBRA 

13.  A  certain  rectangular  field  has  the  same  area  as  another 
which  is  6  rods  longer  and  2  rods  narrower,  and  also  the  same 
area  as  a  third  field  which  is  3  rods  shorter  and  2  rods  wider. 
Find  the  dimensions  of  the  field. 

14.  The  sum  of  the  first  and  third  angles  of  a  certain  tri- 
angle is  twice  the  remaining  angle;  the  sum  of  the  first  and 
second  angles  exceeds  the  third  angle  by  20°.  Find  the  three 
angles  of  the  triangle. 

15.  The  sum  of  the  two  digits  of  a  certain  number  is  14. 
If  36  be  added  to  the  number,  the  result  has  the  same  digits 
as  the  original  number,  but  in  reverse  order.  Find  the  num- 
ber. 

16.  Two  trains,  starting  from  points  270  miles  apart,  and 
traveling  toward  each  other,  will  meet  at  12  o'clock,  if  one 
train  starts  at  7  a.m.  and  the  other  at  10  a.m.  The  rate  of  the 
first  train  exceeds  the  rate  of  the  second  train  by  5  miles  an 
hour.     Find  the  rates  of  the  trains. 

17.  A  boy  can  row  10  miles  downstream  on  a  river  in  two 
hours,  and  can  return  in  3|  hours.  Find  the  rate  at  which  he 
rows  in  still  water  and  also  the  rate  of  the  current  of  the 
river. 

18.  A  train  leaves  A  for  B,  120  miles  distant,  at  9  a.m.,  and, 
one  hour  later,  a  train  leaves  B  for  A.  They  meet  at  noon. 
If  the  second  train  had  started  at  9  a.m.  and  the  first  at  10.30 
a.m.,  they  would  still  have  met  at  noon.     Find  their  rates. 

19.  The  circumference  of  the  hind  wheel  of  a  carriage  is  55 
inches  more  than  that  of  the  fore  wheel.  The  former  makes 
as  many  revolutions  in  going  250  feet  as  the  latter  in  going  140 
feet.     Find  the  circumference  of  each  wheel. 

20.  A  man  has  quarters,  dimes,  and  nickels  to  the  value  of 
$  1.40,  having  in  all  12  coins.  If  he  had  as  man y  dimes  as  he 
has  quarters,  and  as  many  quarters  as  he  had  dimes,  the  value 
of  the  coins  would  be  S1.55.  How  many  coins  of  each  kind 
has  he  ? 


QUADRATIC    EQUATIONS  61 

SUPPLEMENTARY  PROBLEMS 

21.  The  hundreds'  digit  of  a  certain  number  of  three  figures 
is  f  of  the  tens'  digit,  and  exceeds  the  units'  digit  by  2.  If  the 
number  be  divided  by  the  sum  of  its  digits,  the  quotient  is  38. 
Find  the  number. 

22.  r  years  ago,  A  was  m  times  as  old  as  B.  In  s  years,  A 
will  be  7i  times  as  old  as  B. 

(a)  What  are  their  present  ages  ? 

(b)  Find  the  values  of  their  present  ages  if  r  is  10,  s  is  5,  m 
is  5,  and  n  is  2. 

23.  A  man  has  $  14,250  invested  in  bonds,  which  give  him 
annually  a  total  income  of  $700.  Part  of  the  money  is  in- 
vested in  4  %  bonds,  bought  at  $  90  per  share,  and  the  balance 
in  6  %  bonds,  bought  at  $  105  per  share.  How  much  has  he 
invested  in  each  way  ?  (The  income  is  always  computed  on  the 
par  value  of  a  bond,  which  in  this  example  is  $  100  per  share.) 

24.  A  vessel  contains  a  mixture  of  wine  and  water.  If  50 
gallons  of  wine  be  added,  there  will  then  be  J  as  much  wine  as 
water ;  if  50  gallons  of  water  be  added,  there  will  be  4  times 
as  much  water  as  wine.  Find  the  number  of  gallons  of  water 
and  of  wine  at  first. 

25.  The  chords  AB  and  CD  of  a  circle  form,  at  their  inter- 
section, an  angle  of  60°.  The  chords  AD  and  BC,  extended, 
meet  at  0,  forming  an  angle  of  40°.  Find  the  number  of  de- 
grees in  the  arcs  AC  and  DB. 

26.  The  formula  I  =  a  +  (n  —  1)  d  occurs  in  a  more  advanced 
topic  in  algebra.  If  I  is  32  when  n  is  10,  and  is  10  when  n  is 
20,  find  a  and  d. 

27.  The  numbers  d,  a,  t,  and  b  are  assumed  to  be  connected 
by  the  formula  d  =  at  +  b.  If  d  =  f  when  t  =  J,  and  d  =  i| 
when  t  —  | ,  find  a  and  b. 

From  the  resulting  formula  for  d,  determine  t  when  d  is  f , 
giving  the  result  to  the  nearest  eighth  of  an  inch. 


62  ALGEBRA 

28.  Assuming  that  the  numbers  a,  b,d,  and  W  are  connected 
by  the  formula  W=ad-\-b,  find  a  and  b  if  W  =  1.5  when 
d  =  .75,  and  if  TF=  4.5  when  d  =  2.5.  From  the  resulting 
equation,  determine  IF  when  d  =  2. 

29.  If  a  field  were  made  a  feet  longer  and  6  feet  wider,  its 
area  would  be  increased  by  m  square  feet ;  if  its  length  were 
made  c  feet  less,  and  its  width  d  feet  less,  its  area  would  be  de- 
creased by  n  square  feet.     Find  its  dimensions. 

30.  An  automobile  made  a  trip  of  145  miles  in  8  hours.  The 
average  rate  within  city  limits  was  15  miles  an  hour  ;  the  aver- 
age rate  outside  of  city  limits  was  20  miles  an  hour.  Find  the 
part  of  the  trip  lying  within  city  limits  and  the  part  outside. 

31.  In  round  numbers,  the  average  rate  of  the  automobile 
that  won  a  certain  long  auto  race  is  21  times  the  rate  of  an 
ordinary  passenger  train.  At  these  rates,  the  automobile  can  go 
150  miles  in  2  hours  less  time  than  the  train  requires  for  a  trip 
of  140  miles.     Find  the  rate  of  the  automobile  and  of  the  train. 

32.  A  piece  of  work  can  be  done  by  A  and  B  working  to- 
gether in  10  days.  After  working  together  for  7  days,  A 
leaves,  and  B  finishes  the  work  in  9  days.  How  long  would  A 
alone  take  to  do  the  work  ? 

33.  A  motor  boat  which  can  run  at  the  rate  of  r  miles  an 
hour  in  still  water,  went  downstream  a  certain  distance  in 
n  hours  ;  it  took  m  hours  to  return. 

(a)  Find  the  distance  and  also  the  rate  of  the  current. 
(&)  Find  the  values  of  the  two  results  in  part  (a)  when  r  is  10, 
m  is  3,  and  n  is  2. 

34.  A  and  B  can  complete  a  certain  piece  of  work  if  A  works 
5  days  and  B  works  4  days  at  their  usual  rates.  A  and  C  can 
do  the  work  if  they  work  together  for  5  days,  and  if  then  C 
works  one  day  alone.  The  number  of  days  it  would  take  C  to 
do  the  work  exceeds  by  4  days  the  number  required  by  B. 
Find  how  many  days  it  would  take  each  alone  to  do  the  work. 


VII.     SQUARE   ROOT  AND   QUADRATIC   SURDS 

58.  A  Square  Root  of  a  given  number  is  a  number  whose 
square  equals  the  given  number. 

59.  Two  square  roots  are  obtained  for  each  number.  They 
are  of  equal  absolute  value,  but  have  opposite  signs  ;  they  are  in- 
dicated by  means  of  the  double  sign,  ±,  read  "  plus  or  minus." 


Example  1.     Vl6  xiy2  =  ±  4  x*y,  since  (  ±  4  x2y)2  =  16  x4y2. 


Example  2.     V4a?  -  20xy  +  25y2  =  ±(2x-  5y), 

since   \  ±(2x-5y)\2  =  +  (2  x-5  y)2  =  Ax2-  20xy  +  25y2. 

The  positive  square  root  is  called  the  principal  square  root  of 
a  number ;  the  square  root  refers  always  to  the  positive  root. 

60.    The  square  roots  of  a  large  number  may  sometimes  be 
found  by  inspection  by  factoring  the  number. 


Example.     V1764  «4  =  V4  .  441  a4  =  ±  2  .  21  a2  =  ±  42  a2. 

EXERCISE   22 

Find  the  square  roots  of : 

1.    25  a4.  4      16  a6.  6     144^V# 

25  rW  '    81  mV 

2-    ***»•  5     laW.  7      121^o 

3.   49c4d8.  "    49 mV*  '    169  yV* 

8.  If  a  monomial  is  a  perfect  square,  what  kind  of  numbers 
are  the  exponents  of  its  prime  factors  ?    What  sign  does  it  have  ? 

9.  When  is  a  trinomial  a  perfect  square  ? 

10.   How  may  the  correctness  of  the  square  root  of  a  number 
be  checked  ? 

63 


64  ALGEBRA 

Find  the  values  of : 


11.    Vtf2  +  2  xy1  +  y4.  13.    Vl44  x6  -  24  x3?/  +  f 


12.    Va4-6a26  +  9&2.  14.    V49  c4  -  42  c2d2  +  9  d\ 


15.    V1225.  17.    V676a22/2.  19.    V17t>4ary. 


a 

+  b 

a2  +  2  a&  +  &2 
a2 

2a 

+  b 
2a  +  6 

+  2  a&  +  62 
+  2  a6  +  62 

16.    V784m4w2.  18.    Vl089a4.  20.    V1024  xyz2. 

61.  Square  Root  found  by  Long  Division.  If  it  is  not  possible 
to  factor  readily  the  number  under  the  radical  sign,  the  square 
root,  if  there  is  one,  may  be  found  by  a  process  like  long 
division. 

Example  1.     Find  the  square  roots  of  a2  +  2  ah  -j-  62. 

Solution  :  1.    \Ta2  =  a.     Place  a  in  the  root. 

2.  Square  a  ;  subtract. 

3.  2  x  a  =  2  a.     Trial  divisor. 
2  ab  -f-  2  a  =  b.     Add  6  to  the  trial  divisor  and  to 

the  root.     Complete  divisor. 

4.  b  x  (2  a  +  b) ;  subtract. 
The  square  roots  are  ;  +(«  +  &)  and  —  (a  +  b). 

Explanation  :  1.  Find  the  square  root  of  the  first  term,  obtaining  a, 
the  first  term  of  the  root ;  place  it  in  the  root. 

2.  Square  the  first  term  of  the  root  and  subtract  it  from  the  given 
number,  obtaining  the  first  remainder,  2  ab  +  b'2. 

3.  Double  the  first  term  of  the  root,  obtaining  2  a,  the  trial  divisor. 
Divide  the  first  term  of  the  remainder  by  2  a,  obtaining  b,  the  second  term 
of  the  root.  Add  b  to  the  root  and  to  the  trial  divisor  ;  the  complete 
divisor  is  2  a  +  b. 

4.  Multiply  the  complete  divisor  by  b  and  subtract. 

Step  3  is  suggested  by  the  process  of  squaring  a  binomial.  When 
squaring  a  binomial,  the  middle  term  is  obtained  by  taking  twice  the 
product  of  the  first  and  second  terms  ;  this  is  equivalent  to  taking  twice 
the  first  term  and  multiplying  by  the  second.  Reversing  the  process,  the 
second  term,  6,  will  be  found,  if  2  ab  is  divided  by  2  a.  After  a2  is  sub- 
tracted from  a2  +  2  ab  +  b'2  the  remainder  2  ab  +  b2  equals  6(2  a  +  b). 
This  suggests  adding  b  to  the  trial  divisor  and  multiplying  the  sum  by  b. 


SQUARE   ROOT   AND   QUADRATIC   SURDS  6b 

Example  2.     Find  the  square  roots  of 

20  a,-3  -  70  a?  +  4  a4  +  49  -  3  a2. 


Solution  :  1.   Arrange  it 

in  descendir 
Subtract. 

-7). 

lg  powers  of  x  : 
2  x2  +  5  x 

-7 

2.  V4x4  =  2x2. 

3.  (2  x2)2  =  4  x4. 

4x4+20x3-  3x2-70x+49 
4x4 

4.  2  x  (2  &)  =  4  x2. 
20  x3  +  4  x2  =  5  x. 

5.  5x(4x2  +  5  a). 

4x'< 
4x' 

+  5x 
J+5x 

20  x3-  3x2-70x+49 
20x3  +  25x2 

6.  2  x  (2x2  +  6«). 

7.  -28x2-4x2=-7. 

4x2+10x 

-7 

—  28  a? -70  a; +  49 

8.              -7(4x2+10x- 

4x 

2+10 

z-1 

-28x2-70x+49 

The  square  roots  are  :  +  (2  x2  +  5  x  —  7)  and  —  (2  x2  +  5  x  —  7). 

Rule.  —  To  find  the  square  root  of  an  algebraic  expression : 

1.  Arrange  it  according  to  ascending  or  descending  powers  of 
some  letter. 

2.  Write  the  positive  square  root  of  the  first  term  of  the  given 
expression  as  the  first  term  of  the  root.  Square  it  and  subtract 
the  result  from  the  given  expression. 

3.  Double  the  root  already  found,  for  the  trial  divisor.  Divide 
the  first  term  of  the  remainder  by  the  first  term  of  the  trial  divisor. 
Add  the  quotient  to  the  root  and  also  to  the  trial  divisor,  obtaining 
the  complete  divisor. 

4.  Multiply  the  complete  divisor  by  the  new  term  in  the  square 
root ;  subtract  the  product  from  the  remainder  obtained  in  step  2. 

5.  Continue  in  this  manner :  (a)  double  the  root  already  found 
for  a  new  trial  divisor ;  (b)  divide  the  first  term  of  the  remainder 
by  the  first  term  of  this  product  for  the  new  term  of  the  root ; 
(c)  add  the  new  term  of  the  root  to  the  trial  divisor,  obtaining  the 
complete  divisor;  (d)  multiply  the  complete  divisor  by  the  new 
term  of  the  root ;  (e)  subtract. 


66  ALGEBRA 

EXERCISE   23 
Find  the  square  roots  of  the  following : 

1.  25x2-±0xy  +  16y2.  3.    xA  +  6x?  +  11  x2  +  6  x  +  1. 

2.  36  c4  -  60  c2d  +  25  d2.  4.    a4  +  4  a3  +  6  a2  +  4  a  +  1. 

5.  9  n4  +  12  ?i3  -  20  »*  -  16  w  +  16. 

6.  a?2  +  ?/  +  4  z2  —  2  #?/  +  4  #z  —  4  yz. 

7.  8a3-4a-16a4  +  l  +  16a6  +  4a2. 

8.  12  n  -  42  n3  +  4  -  19  n2  +  49  »*. 

9.  a6-2a5-a4  +  6a3-3a2-4a  +  4. 

10.   4z2  +  20a;  +  29  +  — +  i. 

£         iC2 

2      2  aft  ,  13  62     4  63  ,  4  ft4 
a  "3    +     9         9a      9a2' 

12.   f»«-.|:n*-.#»«  +  f*  +  ff 

13     0*   ,    «*    |    3  a?         a;  1 

16     4y      20y*     5tf      25^ 

Find  the  fourth  roots  of : 

14.  a8  -  16  a663  +  96  a456  -  256  a2b9  +  256  612. 

15.  81  a8  -  108  a7  +  162  a6  -  120  a5  +  91  a4  -  40  a3  +  18  a2 

-  4  a  +  1. 
Find. the  first  four  terms  of  the  square  roots  of: 

16.  1  +  aj.  17.    1  —  x.  18.    9  — 2ft 

62.  Square  Root  of  an  Arithmetical  Number.  The  square  root 
of  100  is  10;  of  10,000  is  100;  etc.  Hence,  the  square  root  of 
a  number  between  1  and  100  is  between  1  and  10 ;  the  square 
root  of  a  number  between  100  and  10,000  is  between  10  and 
100;  etc. 


SQUARE   ROOT   AND   QUADRATIC   SURDS  67 

That  is,  the  integral  part  of  the  square  root  of  a  number  of 
one  or  two  figures  contains  one  figure ;  of  a  number  of  three  or 
four  figures,  contains  two  figures ;  and  so  on. 

Hence,  if  the  given  number  is  divided  into  groups  of  two 
figures  each,  beginning  with  the  units  figure,  for  each  group 
in  the  number  there  will  be  one  figure  in  the  square  root.  The 
groups  are  called  Periods. 

Thus,  2345  becomes  23  45  ;  it  has  two  periods  and  its  square  root  has 
two  figures,  a  tens'  and  a  units'  figure. 

34038  becomes  3  40  38  ;  it  has  three  periods  and  its  square  root  has 
three  figures.  A  number  having  an  odd  number  of  figures  will  always 
have  only  one  figure  in  its  left-hand  period,  as  in  this  case. 

A  decimal  number  is  divided  in  the  same  manner,  starting 
from  the  decimal  point  in  both  directions. 

Thus,  3257.846  becomes  32  57.84  60.  The  last  decimal  period  is  always 
completed  by  annexing  a  zero.  The  square  root  of  this  number  has  two 
figures  before  the  decimal  point  and  two  after  it. 

63.  The  first  figure  of  the  square  root  of  a  number  is  found 
by  inspection ;  the  remaining  figures  are  found  in-  the  same 
manner  as  the  square  root  of  a  polynomial. 

Example  1.    Find  the  square  roots  of  4624. 

Solution.  1.  Divide  4624  into  periods  ;  this  gives  46  24.  There  are 
in  the  square  root  a  tens'  and  a  units'  figure. 

2.  The  tens'  figure  must  be  6  ;  7  is  too  large  for  702  =  4900,  which  is 
more  than  4624. 

3.  The  rest  of  the  square  root  is  found  as  follows  : 
3600  is  the  largest  square  less  than  4600. 
V3600  =  60  ;  place  60  in  the  root. 
Square  60  and  subtract. 
Double  60.     Trial  divisor. 

102  -=-  12  =  8+.     Place  8  in  root  and  add  to  trial  divisor. 
Complete  divisor. 
Multiply  complete  divisor  by  8. 

The  square  roots  are  +  68  and  —  68. 

It  is  customary  to  abbreviate  the  solution  by  omitting  the 
zeros  as  in  the  following  example. 


60  +  8 

46  24 

36  00 

120 

10  24 

8 

128 

10  24 

23.5 

5  52.25 

4 

40 

1  52 

3 

43 

1  29 

460 

23  25 

5 
465 

23  25 

68  ALGEBRA 

Example  2.    Find  the  square  roots  of  552.25. 

Solution.     The  largest  square  less  than  5  is  4  ;  VI  =  \ 
Place  2  in  the  root. 

2x2  =  4;  annex  0.     Trial  divisor. 

15  -4-  4  =  3+  ;  add  3  to  the  trial  divisor. 

Complete  divisor.     Multiply  by  3. 

2  x  23  =  46  ;  annex  0.     Trial  divisor. 

230  -=-  46  =  5+.     Add  5  to  the  trial  divisor. 

Complete  divisor.     Multiply  by  5. 

The  square  roots  are  +  23.5  and  —  23.5. 

Rule.  —  To  find  the  square  root  of  an  arithmetical  number : 

1.  Separate  the  number  into  periods  (§62). 

2.  Find  the  greatest  square  number  in  the  left-hand  period ;  write 
its  positive  square  root  as  the  first  figure  of  the  root ;  subtract  the 
square  of  the  first  root  figure  from  the  left-hand  period,  and  to  the 
result  annex  the  next  period.        1 

3.  Form  the  trial  divisor  by  doubling  the  root  already  found  and 
annexing  zero. 

4.  Divide  the  remainder  by  the  trial  divisor,  omitting  the  last 
figure  of  each.  Annex  the  quotient  to  the  root  already  found;  add 
it  to  the  trial  divisor  for  the  complete  divisor. 

5.  Multiply  the  complete  divisor  by  the  root  figure  last  obtained 
and  subtract  the  product  from  the  remainder. 

6.  If  other  periods  remain,  proceed  as  before,  repeating  steps  3,  4, 
and  5  until  there  is  no  remainder  or  until  the  desired  number  of 
decimal  places  has  been  obtained  for  the  root. 

Note  1.  It  sometimes  happens  that,  on  multiplying  a  complete  divisor 
by  the  figure  of  the  root  last  obtained,  the  product  is  greater  than  the 
remainder.  In  such  cases,  the  figure  of  the  root  last  obtained  is  too 
great,  and  the  next  smaller  integer  must  be  substituted  for  it. 

Note  2.  If  any  figure  of  the  root  is  0,  annex  0  to  the  trial  divisor  and 
annex  to  the  remainder  the  next  period. 


SQUARE   ROOT   AND   QUADRATIC   SURDS 


69 


ExAMPLI 

:3.      Fi 

id  the  square  root 

70.32 

Solution  : 

49  44.90  24 

49 

1400 

44  90 

3 
1403 
14060 

42  09 

2  8124 

2 

14062 

2  8124 

The  square  roots  are  +  70.32  and  —  70.32. 

The  first  trial  divisor  is  140.  Since  this  is  greater  than  44,  the  first 
remainder,  annex  0  to  the  root,  obtaining  70. 

The  second  trial  divisor  is  1400;  (2x70=140;  annex  0,  1400). 
Bring  clown  the  next  period  90,  getting  for  the  second  remainder  4490. 
Divide  44  by  14  gives  3+  ;  annex  3  to  the  root  and  add  3  to  1400,  etc. 


EXERCISE  24 
Find  the  square  roots  of : 

1.  5776.  4.    8427.24.        7.    54.4644.  10.    106  09. 

2.  15376.        5.    7974.49.        8.    1488.4164.        11.    529.9204. 

3.  67081.        6.    11.6281.        9.    25.6036.  12.    1592.8081. 

64.  The  Approximate  Square  Roots  of  a  number  which  is  not 
a  perfect  square  are  often  desired.  Obtain  usually  the  first 
three  figures  following  the  decimal  point. 

Example.     Find  the  approximate  square  roots  of  2. 

Solution  :  1.414 


2.00  0000 

20 

4 

2? 

1 
100 

96 

2* 

JO 

4  00 

2ST 

2  81 

2820 

1  19  00 

2  8"' 

>4 

112  96 

The  square  roots  are  +  1.414+  and 


7  04 
-  1.414 


70  ALGEBRA 

Note.  In  order  to  obtain  the  desired  number  of  decimal  places,  annex 
zeros  until  there  are  three  periods. 

EXERCISE   25 
Find  the  approximate  square  roots  of : 

1.  3.         3.    6.  5.    10.  7.    13.  9.    15.  11.    19. 

2.  5.         4.    7.  6.    11.  8.    14.         10.    17.  12.   21. 

65.  Table  of  Square  Roots.  In  the  remainder  of  the  course, 
it  will  be  necessary  to  use  frequently  the  square  roots  of 
some  numbers.  Retain  some  of  the  square  roots  as  they  are 
found,  either  in  a  notebook  or  in  some  other  convenient  place. 
Make  a  list  of  the  numbers  from  1  to  50,  and  write  their 
square  roots  beside  them,  thus : 

Number  Square  Root 

1  1.000 

2  1.414 

3  .  1.732 

After  working  Exercise  25,  twelve  of  the  numbers  of  this 
table  may  be  tabulated.  These  roots  may  be  used  to  obtain 
the  square  roots  of  other  numbers. 

Example  1.     Find  the  square  roots  of  8. 

Solution  :    V8  =  \'TV2  =  2  x  V2 :  =  2  x  ( ±  1.414+)  =  ±  2.828+. 

Example  2.     Find  the  square  roots  of  12. 

Solution  :  Vl2  =  V4  x  3  =  2V~3  =  2  x  (±  1.732+)=  ±  3.464+. 

EXERCISE  26 

1.  Find  the  following  square  roots  to  three  decimals: 

(a)   Vl8.        (6)  V20.        (c)  V24.         (d)  V27.         (e)  V28. 

2.  Complete  your  table  of  square  roots  up  to  50.  Get  as 
many  roots  as  possible  by  inspection  (§  60) ;  get  as  many 
of  the  remaining  roots  as  possible  as  in  Example  1.  Find  the 
others  by  the  long  division  method  (§  62). 


SQUARE   ROOT   AND   QUADRATIC   SURDS  71 

66.    The  square  roots  of  a  fraction  which  is  not  a  perfect 
square  may  be  found  as  follows  : 


X  2       ^4    '  V4 

V6  =      2^49+ =         >224+> 
^2  2 

Rule.  —  To  find  the  square  root  of  a  fraction : 

1.  Change  the  fraction  into  an  equivalent  fraction  with  a  perfect 
square  denominator. 

2.  The  square  root  of  the  new  fraction  equals  the  square  root  of 
its  numerator  divided  by  the  square  root  of  its  denominator. 

3.  If  desired,  express  the  result  of  step  2  in  simplest  decimal 
form,  prefixing  the  double  sign,  ± . 

Example.     Find  the  approximate  square  roots  of  |. 

Solution  :    1.    The    smallest   square    number   into    which   8   can  be 
changed  is  16 ;  multiply  both  terms  of  the  fraction  by  2. 

%    JI  =  J2Z3  =  ^=±^=±^491=  +< 

>8       ^2  x  8       *  16  4  4 

EXERCISE  27 
Find  the  approximate  square  roots  of : 


1. 

9 
f' 

3. 

ft- 

5. 

4 

7. 

2. 

9. 

5 

11. 

5 

XT' 

2. 

i 

4. 

1 

T 

6. 

5 
2* 

8. 

3 

V 

10. 

h 

12. 

ft- 

QUADRATIC   SURDS 
67.    The  indicated  square  root  of  a  number  which  is  not  a 
perfect  square  is  called  a  Quadratic  Surd;  as,  V8,  «%/->   Vsc, 


^ 


*+i. 


72  ALGEBRA 

68.  Surds  should  be  simplified  as  in  the  following  examples  : 

(a)  V24=V4T6=2.V5;  (6)  ^|  =  ^  =  §^?. 

Thus,  a  quadratic  surd  is  in  its  simplest  form  when  the 
number  under  the  radical  sign  is  an  integer  which  does  not 
contain  any  perfect  square  factor. 

In  problems  involving  surds,  it  is  agreed  to  consider  for 
each  surd  only  its  principal  root  (§  59). 

69.  Addition  and  Subtraction  of  Surds. 
Example  1.     Find  the  sum  of  V20  and  V45. 

Solution  :  1.    V20  +  V45  =  VlTl  +  ViT~5  =  2V5  +  8VS  =  5 VE. 

This  solution  assumes  that  surds  may  be  added  like  other  numbers. 
The  coefficients  of  V5  are  2  and  3  ;  the  sum  is  found  by  multiplying  V5 
by  the  sum  of  its  coefficients  (§  5). 

The  advantage  in  adding  surds  in  this  way  is  that  fewer  square  roots 
need  be  obtained.  Thus,  the  sum  of  V20  and  V45  is  5  V5  or  5  x  (2.23(5+ ) 
or  11.180+.  This  same  result  couid  be  obtained  by  adding  the  square 
roots  of  20  and  45. 

Example  2.     Simplify  Vf  +  VJ. 

0  ,     JO  ,     /l        /18  ,     /2     3V2  .  V2     3V2  .  2V2     5V2 

2     5  V2  Jx  (1.414+)  _  7.070+  _  1  ?67+ 
4  4  4 

Example  3.     Simplify  |  +  VJ. 

k  1      2  ,  Jl      2  ,  ./3      2  ,  V3      2+V3 


2. 


2+V3  =  2  +  1.732+  =  3.732+  =12u+ 


Note.  The  results  of  problems  involving  surds  are  often  left  in  the 
surd  form  as  in  step  1  of  Examples  2  and  3.  There  are  advantages  in 
finding  the  approximate  decimal  value  of  the  result, 


SQUARE   ROOT   AND   QUADRATIC   SURDS  73 


EXERCISE  28 

Simplify  the  following: 

1.    V12  +  V2T. 

11. 

i  +  VA- 

2.    V20-V5. 

12. 

l+VS- 

3.   2V18  +  V98. 

13. 

I+VJ. 

4.    V63-2V28. 

14. 

*-Vf. 

5.   3V24-V54. 

15. 

I-VS- 

6.    2V2  +  VI8- 

V50. 

16. 

-I  +  Vf. 

7.    V8+VJ. 

17. 

-i-vi. 

8.    Vf-Vf 

18. 

1  _a/6 

9.    1  +  V|. 

19. 

11     "^  V  12  1* 

10.  f-Vf. 

20. 

1  1  _|_  a/ 2T 

70.  The  other  operations  with  surds,  namely,  multiplication, 
division,  involution,  and  evolution,  are  considered  in  a  later 
chapter,  which  may,  if  desired,  be  studied  at  this  time. 


VIII.   QUADRATIC   EQUATIONS 

71.  A  Quadratic  Equation  is  an  equation  of  the  second  degree 
(§  45) ;  it  may  have  one  or  more  unknowns. 

A  Pure  Quadratic  Equation  is  a  quadratic  equation  having 
only  one  unknown,  which  contains  only  the  second  power  of 
the  unknown,  as,  ax2  =  b. 

Example  1.  An  acre  of  ground  contains  43,560  square  feet. 
How  long  must  the  side  of  a  square  field  be  in  order  that  the 
area  of  the  field  shall  be  one  acre  ? 

Solution  :  1.    Let  s  =  the  number  of  feet  in  one  side. 

2.  Then  s2  =  the  number  of  square  feet  in  the  area. 

3.  Then  s2  =  43,560. 

Extract  the  square  root  of  both  members  of  the  equation. 

4.  Then  s  =  ±208.7+. 

Since  this  is  a  field,  only  the  positive  root  has  meaning ;  hence  the 
side  of  the  field  must  be  208.7+  feet. 

72.  A  pure  quadratic  equation  has  two  roots,  because  two 
square  roots  are  obtained  in  extracting  the  square  roots  of  the 
two  members  of  the  equation. 

Rule.  —  To  solve  a  pure  quadratic  equation. 

1.  Clear  the  equation  of  fractions,  transpose,  and  combine  terms 
until  the  equation  takes  the  form  x1  =  a  number. 

2.  Extract  the  square  roots  of  both  members  of  the  equation, 
placing  the  double  sign,  ± ,  before  the  root  in  the  right  member. 

Note.  After  extracting  the  square  roots  of  both  members  of  an  equa- 
tion like  x2  =  a2,  we  get  ±  x  =  ±  a.  This  gives  :  +  x  =  +  a,  +  x  =  —  a, 
—  x  =  +  a,  and  —  x  =  —  a. 

If  both  members  of  the  last  two  equations  are  multiplied  by  —  1,  the 
equations  become  +  x  =  -  a,  and  +  x  =  +  a.  These  are  the  first  two 
of  our  four  equations.  Thus,  it  is  clear  that,  from  x2  =  a2,  we  get  only 
two  equations,  x  =  +  a  and  x  =  —  a,  or  x  =  ±  a. 

74 


QUADRATIC   EQUATIONS  75 

Example.     Solve  the  equation 1 —  =  -—-}-  —  . 

3       m      12      m 

0  .,  2  to  .    3       to    .  12 

Solution  :     1. 1 = 1 

3        to      12      to 

2.  M12m:  8  to2  +  36  =  to2  +  144. 

3.  Simplifying  :  7  w2  =  108. 

4.  D7:  m2  =  i^. 

5.  y/~ ":*  w»  =±v^=±6Vf  =  ±fV21. 

6.  V21=  4.582:  to  =  ±  2  .  (4.582)  =  ±  ^|^  =  ±  3.927. 

7.  To!  =+  3.927  ;  to2  =-3.927. 

"  toi  "  is  read  "  to  one."  The  numeral  1  is  called  in  such  cases  a  sub- 
script, "wa"  is  read  "to  two."  These  subscripts  are  used  to  distin- 
guish between  the  two  roots  of  the  quadratic. 

Check  :  When  the  roots  are  complicated,  it  is  better  to  check  by  going 
over  the  solution  a  second  time.  Great  care  must  be  taken,  however,  for 
it  is  easy  to  overlook  an  error. 

Note.  Get  the  result  in  the  radical  form  first ;  that  is,  to  =  i  f  V21 ; 
then  it  is  wise,  for  many  reasons,  to  get  it  in  decimal  form  as  finally  given. 


EXERCISE  29 
Solve  the  following  equations  : 

1.  5  c2 -180  =  0.  3.   13  c2 -135  =  10  c2 -27. 

n       0^2    ,     O-         -      2         KQ*  A      4*2  +  3         8*2-l  1 

2.  2  x2  +  27  =  i  x2  —  53.  4. 


10.     ^  —  -r- 


6.  5(t  +  6)-t(t-3)=$t. 

7.  9a2-5  =  0. 

8.  lla2-6  =  3. 

9     A_J1  =  _?.  ii 

4f      Sx2         3  "    2c  +  l       c-1 

*  The  symbol  "V   ":    placed  in  the  left  margin  will  mean,   "take  the 
square  root  of  both  members  of  the  previous  equation." 


7 

2 

14 

2)  = 

10  m. 

2x 
3  " 

4:X 

Ix 
9 

21 
2x 

5  c- 

-2     3 

c  — 

5 

76  ALGEBRA 

1  1  a2- 17 


12. 


x  -f  3     as  —  5     a;2  —  2  x  —  15 


a?2  —  a;  -|-  2      x2  +  x  —  3  _. 
x—  2  x  -f-3 


14. 


3  a  a?4-5&    _  ^ 


a  -  5  6      3«4-10  6 


15.    a2  -  2  ca2  =  3  b\     Solve  for  a. 

Solution  :  1.  —  2  ex2  =  3  62  -  a2. 

2.  M_i:  2cx2  =  a2-3  62. 

3.  z2  =  ^ 


2c 


__        U-3^  /2c(«*-3fr2) 

\       2  ^  X  \  4  ri* 


±  —  V2  a2c  -  6  62c. 
2c 


PROBLEMS   IN   PHYSICS 


All  of  the  following  equations  occur  in  the  study  of  physics. 
Solve  them  for  the  numbers  which  appear  with  exponent  2. 


;.  S  =  ±gt\ 

18. 

F_mM 
d2 

20.  /=^ 
J       R 

.   E  =  ±mv\ 

19. 

H=  C21U. 

21-    #=ls 

D2 

73.  Geometry  Problems.  If  the  following  terms  from  ge- 
ometry are  not  familiar  to  the  student,  they  should  be  re- 
viewed :  (a)  right  triangle ;  (b)  hypotenuse ;  (c)  isosceles  tri- 
angle ;  (d)  equilateral  triangle ;  (e)  circle ;  (/)  theorem ; 
(g)  altitude  of  a  triangle  ;  (h)  base  of  a  triangle. 


QUADRATIC   EQUATIONS  77 

EXERCISE   30 

Carry  out  all  results  in  this  exercise  to  one  decimal  place : 

1.  State  the  theorem  about  the  square  of  the  hypotenuse  of 
a  right  triangle. 

2.  Find  the  altitude  of  a  right  triangle  whose  base  is  13  feet 
and  whose  hypotenuse  is  30  feet. 

Solution  :  1.  Let  x  =  the  number  of  feet  in  the  altitude. 

2.  Then  x2  +  132  =  302.  (why  ?) 

3.  Complete  the  solution. 

3.  Find  the  base  of  a  right  triangle  whose  hypotenuse  is 
45  feet  and  whose  altitude  is  27  feet. 

4.  If  the  altitude  of  a  rectangle  is  h  feet  and  its  base  is 
four  times  its  altitude,  find  the  length  of  the  diagonal. 

5.  Solve  the  formula  h2  =  o2  +  b2:  (a)  for  a ;  (6)  for  b. 

6.  Find  the  altitude  of  an  isosceles  triangle  whose  equal 
sides  are  each  15  inches  and  whose  base  is  8  inches. 

7.  Find  the  altitude  of  an  isosceles  triangle  if  its  equal 
sides  are  each  4  b  inches  and  its  base  is  2  b  inches. 

8.  Find  the  altitude  of  an  equilateral  triangle  if  its  sides 
are  each  8  inches. 

9.  Find  the  altitude  of  an  equilateral  triangle  if  its  sides 
are  each  a  inches. 

10.  (a)  What  is  the  formula  for  the  area  of  a  circle  ? 
(b)  Find  the  area  of  the  circle  of  radius  7  inches. 

Express  the  results  of  the  following  examples  in  simplest  radical 
form  : 

11.  Solve  the  equation  A  =  -n-r2  for  r:  (a)  letting  ir  =  3j; 
(b)  without  substituting  for  it  its  value. 

12.  The  volume  of  a  circular  cone  is  given  by  the  formula 
V=  \  irr2h,  where  r  is  the  number  of  units  in  the  radius  and  h 
is  the  number  in  the  altitude.  Find  V  when  r  =  5  feet  and 
ft  =  13  feet. 


78  ALGEBRA 

13.  Find  the  radius  of  a  circular  cone  whose  volume  is  528 
cubic  feet  and  whose  altitude  is  14  feet. 

14.  Solve  the  formula  for  the  volume  of  a  circular  cone  for 
r  in  terms  of  V,  h,  and  w. 

15.  Solve  the  formula  8  =  4  irr2  for  r. 

16.  The  distance  s,  in  feet,  through  which  an  object  falls  in 
t  seconds  is  given  by  the  formula  a  =  \  gt2,  where  g  =  32. 

Suppose  that  a  stone  is  allowed  to  fall  from  a  tower;  how 
far  will  it  fall  in  :  (a)  3  seconds  ?  (b)  5  seconds  ? 

17.  How  long  will  it  take  a  ball  to  fall  300  feet  ? 

18.  Washington's  Monument  in  Washington,  D.C.,  is  555  feet 
high.     How  long  will  it  take  a  ball  to  fall  that  distance  ? 

19.  Solve  the  formula  F==  2  ir2Rr2  for  r. 

20.  Solve  the  formula  v .=  J  irr2a  for  a. 

COMPLETE  QUADRATIC  EQUATIONS 

74.  A  Complete  Quadratic  Equation  is  a  quadratic  equation 
having  only  one  unknown,  which  contains  the  first  power  of 
the  unknown  as  well  as  the  second  power ;  as, 

2a2-3a-5:=0. 

A  complete  quadratic  equation  may  be  Solved  by  Factoring. 
The  solution  is  based  upon  the  fact  that  if  one  of  the  factors  of 
a  product  is  zero,  the  value  of  the  product  is  also  zero. 

Thus,  3x0  =  0;   (-5)  x0  =  0;  2x0x  (-  3)  =  0  x  (-3)=0. 

Example  1.     Solve  the  equation  4  x2  —  9  =  0. 

Solution:  1.    Factor:  (2se— 3)(2as  +  3)  =,0. 

2.  If  2  x  -  3  =  0,  then  (2  x  -  3)  (2  x  +  3)  =  0. 

2x—  3=0,  if  2 ac  =  3  or  a;  =  +  |. 

3.  If  2  x  +  3  =  0,  then  (2  x  -  3)  (2  x  +  3)  =  0. 

2  x  +  3  =  0,  if  2  x  =  -  3,  or  x  =  -  f .      . 

4.  The  roots  of  the  equation  are  +  f  and  —  f . 


QUADRATIC   EQUATIONS  79 

5.    Check:  Does  4(|)2_9  =  0? 

1    8 

Does  jl  •  -  -  9  =  0  ?  i.e.  9-9  =  0?    Yes. 
i 

Does  4(- |)2 -9  =  0? 

1    9 
Does  4- --9  =  0?  i.e.  9-9  =  0?     Yes. 

i 

Rule.  —  To  solve  an  equation  by  factoring : 

1.  Transpose  all  terms  to  the  left  member. 

2.  Factor  the  left  member  completely. 

3.  Set  each  factor  equal  to  zero,  and  solve  the  resulting  equations. 

4.  The  roots  obtained  in  step  3  are  the  roots  of  the  given  equa- 
tion.    Check  by  substitution  in  the  given  equation. 

Example  2.     Solve  the  equation =  — 

H  3       2       6 

Solution  :  1.  M6 :  *  2  m2  —  3  m  =  35. 

2.    S35  :  2  m2  -  3  m  -  35  =  0. 

■    3.    Factor  :  (2  m  +  7)  (w  -  5)  =  0. 

4.  2  m  +  7  =0,  if  ra=-f. 

m  —  5  =  0,  if  m  =  +  5. 

5.  The  roots  of  the  equation  are  +  5  and  —  |. 
Check  by  substitution. 

EXERCISE  31 

Solve  the  following  equations  by  factoring : 

1.  x2-15x  +  51  =  0.  5.    8a2-10a:-r-3  =  0. 

2.  tf  +  4a«96.  6    x2  +  7x  =  Q. 

3.  .t2=^  +  H0.  7<    x*  +  ax-2a2  =  0. 

4.  6  x2  +  7  x  +  2  =  0.  (Solve  for  x.) 

*  For  meaning  of  "  M6  :  "  see  §  36. 


80  ALGEBRA 


8.    3z2-  mz-4m2  =  0.  1E     5-t        12 

15. 


9.    15x2  +  xk  =  2W. 
10.    10x2  +  7mx  =  12m2. 


16. 


11.    ^!_?_?_*?  =  o.  17. 

10       5       2 


12.    ?-X«X.  18: 


3 

—  t 

6- 

-£ 

s 

8 
-3 

s- 

if*-": 

V 

5 
-3 

y- 

3         1 
-4      6 

V 

2 

-3 

l 

6 

6 

p  —  S 

3      9#     3z2 


13     A__l  =  1.  19    3      2Q+6)  =     ro  +  6^ 

'    2x2     4  a     4*  m+5  2 


14.    5-lji*.  20.    ^  =  -5 8^- 

6     2     »as  *  — 2     8      (z-2)2 

75.  Graphical  Solution  of  Equations  with  One  Variable.  Many 
facts  about  equations  containing  one  variable  can  be  discovered 
by  the  aid  of  graphical  representation. 

Example  1.     Consider  the  equation  Sx  —  12  =  0. 

The  expression  3  x  —  12  has  a  different  value  for  each  value  of  x. 
Thus,  if  x  =  2,  3  x  -  12  =  -  6  ;  if  X  =  -  3,  3  x  -  12  =  -  21. 
The  problem  is  to  find  the  value  of  x  for  which  the  expression  3  x  —  12 
will  equal  zero. 

Graphical  Solution  :  1.    Let  y  =  3  x  —  12. 

2.  Find  values  of  y  for  some  values  of  x  : 

if  x  =  0,  y=-\2;  tf  x  =  -2,  y  =  -  18  ; 

if  a;  =  +  5,  */  =  +  3  ;  if  x  =  +  6,  y  =  +  6. 

3.  Use  these  pairs  of  numbers  as  coordinates  of  points  and  draw  the 
graph. 


QUADRATIC    EQUATIONS 


81 


Y 

i 

r 

/ 

/ 

/ 

/ 

/ 

L 

; 

/ 

^ 

* 

d 

A/ 

X 

-1 

p 

10 

. 

/ 

u/ 

_J 

Br 

y 

4.  ^C  crosses  the  x  axis  at  point  A.    The  coordinates  of  A  are  :  a;  =  4, 

y  =  o. 

5.  Hence  when  x  =  4,  3  x  —  12  =  0.     (y  is  the  expression  3  x  —  12.) 

.'.  x  =  4  is  the  desired  solution  of  the  equation,  for  we  were  looking  for 
a  value  of  x  for  which  3  x  —  12  =  0. 


Rule.  —  To  solve  graphically  an  equation  containing  one  variable : 

1.  Simplify  the  equation  as  much  as  possible. 

2.  Transpose  all  terms  to  the  left  member. 

3.  Represent  by  y  the  expression  found  in  step  2. 

4.  Find  for  y  the  values  which  correspond  to  selected  values  of 
the  variable  in  the  equation. 

5.  Use  the  pairs  of  values  obtained  in  step  4  as  coordinates  of 
points ;  plot  the  points ;  draw  the  graph,  making  the  vertical  axis 
the  y  axis. 

6.  The  graph  crosses  the  horizontal  axis  at  points  whose  ordi- 
nates  are  zero,  and  whose  abscissae  are  the  desired  roots  of  the 
equation. 


82 


ALGEBRA 


Example  2.     Solve  the  equation  x2  —  x  =  6. 

Solution  :     1 .   x2  —  x  =  6,  or  x2  —  x  —  6  =  0. 

2.  Let  y  =  £2  —  ae  —  6. 

3.  If  x=-4,  2/=(_4)2-(-4)-6  =  16  +  4-6  =+14. 
4. 


Similarly  if  x  = 

0 

+  1 

+  2 

+  4 

+  5 

-  1 

-2 

-3 

-  4 

then  ?/  = 

-6 

—  6 

-4 

+  6 

+  14 

-4 

0 

+  6 

+  14 

\              y-                n 

4                         3-        Lit 

t                               J 

\                                   r 

- '                +10                   ""'  '  j  ' '  ■■ 

"_A_                                      _J             it 

4                                1-7 

3                               r 

V       -a,   .     .  _L__jr      in 

\                           / 

>                           t 

\                         /! 

Xi'                                  \A       j     O               B/                            X 

-7     -6    -5     -4     -3    -2\    -               +1    |+?     +5    +4    +5    +6    +7 

JL                                                                                    J 

\                                                                             / 

V^                                            /l 

>.     y 

*"***  • 

j 

_J_y£ L 

5.  The  graph  crosses  the  horizontal  axis  at  the  points  A  and  B.  Ac- 
cording to  the  rule,  the  abscissae  of  these  points  are  the  two  roots  of  the 
equation. 

At  A:  a:  =  -  2,  y  =  0  ;  i.e.  x2  -  x  -  6  =  0. 

At  B:  a;  =  +  3,  y  =  0;  i.e.  x2  -  x  -  6  =  0. 

Check:        z  =  -  2  ;  does  (-  2)2  -  (-  2)-  6  =  0  ?     Yes. 
x  =  +  3;  does  (+3)2-(+  3)- 6  =  0?     Yes. 


EXERCISE  32 

Solve  graphically  the  equations  : 

1.  #-3  =  0.  3.   x2-9  =  0. 

2.  2x  =  9.  4.    x2-\-3x  =  10. 


5.  a2-7a,'  +  10  =  0. 

6.  a2 +  7  a; +  6  =  0. 


QUADRATIC   EQUATIONS  83 

7.  Between  what  two  integers  does  each  of  the  roots  of  the 
following  equation  lie  ?     4  x2  —  4  x  —  35  =  0. 

Obtain  the  approximate  roots  of  the  following  equations  to 
the  first  decimal  place. 

8.  2  x2  -  x  -  11  =  0.  9.   x2  +  3  x  -  14  =  0. 

76.    Solution  by  Completing  the  Square. 
Development  1.     Find:  (a)  (x  —  4)2;  (b)  (#  +  5)2; 

2.  When  is  a  trinomial  a  perfect  square  ?     (See  §  15,  c.) 

3.  Make  a  perfect  square  trinomial  of  x2  — 10  x. 

Solution  :  1.    |  of  10  =  5  ;  52  =  25  ;  add  25. 

2.    The  perfect  square  is  x2  —  10  x  -f  25  or  (x  —  5)2. 

t    4.    Make  perfect  square  trinomials  of  the  following : 
(a)  a2-12z;  (6)  y2-Uy,  (c)  z2  -  20  2. 

5.  Solve  the  equation  x2  — 12  x  +  20  =  0. 

Solution  :  1.  S20  :   «2  —  12  £  =  —  20. 

2.  Make  the  left  member  a  perfect  square  by  adding  36  ;  therefore 
add  36  to  both  members  (§§  35,  37). 

A36  :  x2  -  12  x  +  36  =  36  -  20, 

or  (x- 6)2  =  16. 

3.  V":  z-6=±4. 

4.  .-.  x  —  6  =  +  4,  or  x  =  6  +  4  =  10,  one  root, 
and  x  —  6  :=  —  4,  or  x  —  Q  —  4=2,  another  root. 

Check  :  x  =  10  ;  does  (10)2  -  12(10)  +  20  =  0  ?     Yes. 

x  =  2  ;  does  (2)2  -  12(2)  +  20  =  0  ?    Yes. 

6.  Solve  the  equation  x2  —  3  #  —  5  =  0. 

Solution  :  1.   x'1  —  3  x  —  5  =  0. 

2.  A6:       x2-3x  =  +  5. 

3.  £(-  3)  =  -  |  ;  (-  $)2  =  +  | ;  add  f  to  both  members. 

4.  A9:       x2-3x  +  f  =  5  +  f  =  -«£. 


84  ALGEBRA 

5.    V~:  x  -f  =±  V^  =  ±^V29. 

2      2  2 

7.  Radical  results,     XX  =  3  +  ^  and  x2  =  S~^29 . 

2  2 

8.  Decimal  results,    xx  =  S  +  5'385  and  a;2  =  8  ~  f'385 

2  2 

8.385  _  -  2.885 

2  '  2 

=  4.192-^  =-1.192+ 

Check  :  To  check  the  solution  by  substituting  the  roots  in  either  their 
decimal  or  their  radical  form  is  a  long  process,  with  many  opportunities 
for  errors.  Persons  skillful  in  algebra  check  by  going  over  the  solution 
carefully. 

A  quick  check,  the  reason  for  which  will  be  learned  later  in  algebra,  is 
to  find  the  algebraic  sum  of  the  roots ;  this  result  should  equal  the  nega* 
tive  of  the  algebraic  coefficient  of  x  in  the  equation  in  which  the  coefficient 
of  x2  is  1. 

Here  :  +  4.192+  The  coefficient  of  x2  is  1.     The  coefficient  of  x 

—  1.192+       is  —  3.     This  equals  the  negative  of  the  algebraic 

Sum.     +  3  sum  of  the  roots. 

If  the  coefficient  of  x2  is  not  1,  first  imagine  the  equation  divided  by 
that  coefficient,  and  then  select  the  coefficient  of  x. 

Rule.  —  To  solve  a  quadratic  equation  by  completing  the  square  i 

1.  Simplify  the  equation ;  transpose  all  terms  containing  the 
unknown  number  to  the  left  member,  and  all  other  terms  to  the 
right  member  so  that  the  equation  takes  the  form 

ax2  -f-  bx  =  c 

2.  If  the  coefficient  of  x2  is  not  i,  divide  both  members  of  the 
equation  by  it,  so  that  the  equation  takes  the  form 

x*+px=q. 

3.  Find  one  half  of  the  coefficient  of  x,  square  the  result ;  add  the 
square  to  both  members  of  the  equation  obtained  in  step  2.  This 
makes  the  left  member  a  perfect  square. 


QUADRATIC   EQUATIONS  85 

4.  Write  the  left  member  as  the  square  of  a  binomial;  express 
the  right  member  in  its  simplest  form. 

5.  Take  the  square  root  of  both  members,  writing  the  double 
sign,  ± ,  before  the  square  root  in  the  right  member. 

6.  Set  the  left  square  root  equal  to  the  +  root  in  the  right 
member  of  the  equation  in  step  5.  Solve  for  the  unknown.  This 
gives  one  root. 

7.  Repeat  the  process,  using  the  -  root  in  step  5.  This  gives 
the  second  root  of  the  equation. 

8.  Express  the  roots  first  in  simplest  radical  form,  and  then,  if 
desired,  in  simplest  decimal  form. 

EXERCISE  33 

Solve  by  completing  the  square: 

1.  x2  +  4  x-  5  =  0.  10.  ra2-fl0m  =  3. 

2.  ^2_8a,_33  =  o.  11.  x2  +  3#-4  =  0. 

3.  ^ +  6^-27  =  0.  12.  s2  =  5s  +  6\ 

4.  x2  -4- 10^  +  21  =  0.  13.  y2  +  3y  =  10. 

5.  a2  -  12  x- 13  =  0.  14.  z2  +  z  =  6. 

6.  y2-2y  =  \l.  15.  r2-3  =  r. 

7.  «2  +  6a  =  9.  16.  z2  =  2  +  3z. 

8.  c2-4c  =  l.  17.  w2-f-5w;  +  3  =  0. 

9.  d2-8d-8  =  0.  18.  a2-7a  +  7  =  0. 

19.    Solve  the  equation  x2  —  %x  =  1. 

Solution  :  1.  \  of  (-  f)  =  -  \  ;  (-  i)2  =  \. 

2.  A^:  x2-fx  +  i  =  l  +  i 

3.  (x  -  |)«  =  V». 

4.  x-\=±^T0. 


86 

ALGEBRA 

5.  x-l  =  l VlO. 

*-f 

=  -iVTo. 

.•.*  =  |  +  §v1o 

X 

=  i-iVlo 

_1+Vl0 

1-  VlO 

3 

3 

_  1  +  3.162 

_  1  -  3.162 

3 

3 

=  4-162  =  1.387+. 
3 

_  -  2.162  _       ? 
3 

20. 

*  +  {«..  |. 

33. 

i_  A_2. 

21. 

2/2-f2/  =  5. 

3x     x2' 

22. 
23. 

22  _   6  f  _  1  =  0. 

a2  +  ia  =  f. 

34. 

2x-\--  — 

2     4a; 

24. 

h2  -  3  &  _  5  =  0. 

35. 

2/  ,  3_        2 

25. 

/2  _   5  *  _  2  5 
fc            6"  fc  —  ~6- 

3     2          32/ 

1  .    3  _.   5 

5     4a     4a2 

26. 

3a2-2a;  =  40. 

36. 

27.  4  m2  —  8ra  =  45. 

o*       24        24      , 

28.  8r2+2r=3.  3?-    - — 5 =  1- 


31.   9c2-f-18c  =  -8. 


a:  —  z       a; 


29.  4*2-3£  =  3. 

38        5      +     8     -3 

30.  a;2  +  7a;  =  5.  38'    5Z^  +  83p-* 


39. 


d-S      d+4      3 


32.    9#2  +  4a;=6.  d-2         d         2 

77.    Solution  of  Literal  Quadratic  Equations. 
Example.     Solve  the  equation  ax2  —  3  bx  —  c  =  0. 
Solution  :  1.  a#2  -  3  bx  —  c  =  0. 

2.  D0:  X2_§_^_c=0 

a         a 

3.  Ac:  ^-Ma^i. 

«  a         a 


tion 


QUADRATIC   EQUATIONS  87 

4.   The  coefficient  of  x  is  (z1^) ;  one  half  of  it  is  f  ~^-\ 

The  square  of  f  —        )  is  ( ].     Add  this  to  both  members  of  equa- 


5.  x2_3_^+^  =  ^  +  c 

a         4  a'2     4  a2     a 


\('-K) 


3ft\2     9  62  +  4ac 


4  a2 
,    1 


-  =  ±  -±-  V  9  ft2  +  4  ac. 
2a  2a 


_  +  3  6  ±  V9  ft2  +  4  ac 
2a 


Q      .        _  +  3  ft  +  V9  ft2  +  4  ac .     8  _  +  3  b  -  V9  ft2  +  4  ac 
2a  2a 

Check  :  xx  +  x2  =  — — -  =  H — -.     Since  this  is  the  negative  of  the 
2a  a 

coefficient  of  x  in  step  2,  the  roots  are  correct. 

EXERCISE   34 

Solve  the  following  equations  for  x : 

1.  x2  +  2mx  =  l  —  m2.  8.   a2  —  2a#  =  9  —  6  a. 

2.  x2  +  6ax-5  =  0.  9.   z2-10ta  =  -9*2. 

3.  a,2-2ax  +  6=0.  10.    ax2  +  4  #  +  1  =  0. 

4.  a2  +  6#-c=0.  11.    to2  +  2  ex- 3  =  0. 

5.  #2+|>a;+  g  =  0.  12.    cx2  + 2  d#  + e  =  0. 

6.  2a2  +  6a     n  =  0.  13.    ax2  +  bx  =  0. 

7.  2  a:2  +  4  ax  —  c  =  0.  14.    ax2  +  bx  +  c  =  0. 

78.    Solution  of  Quadratic  Equations  by  a  Formula.     All  quad- 
ratic equations  having  one  unknown  may  be  put  in  the  form 

ax2+bx  +  c  =  0. 


88  ALGEBRA 

This  equation  has  been  solved  as  Example  14  of  Exercise  34. 

The  roots  are  :  ,         ITz -. — - 

—  b  ±  V62  —  4  ac 

2a 

This  result  is  used  as  &  formula  for  solving  any  quadratic 

equation  of  the  form  ax2  +  bx  -f-  c  =  0. 

Example  1.     Solve  the  equation  2  x2  —  3  x  —  5  =  0. 
Solution  :     1.    Comparing  the  equation  with  ax2  +  bx  -f  c  =  0: 

a  =  2,  b  =—  3,  c  =  —  5. 
2.    Substitute  these  values  in  the  formula : 


3.   Then 


-  b  ±  Vb2  -  4  ac 
2a 

-(-3)±V(-3)2- 

-4(2)(- 

■5) 

2(2) 

+  3  ±  V9  +  40 
4 

3  ±  V49  _  3  ±  7 
4                4 

3  +  7_10      5 

4      -4-2'^- 

_3-7_ 
4 

^i=-l. 
4 

4.  /.  xi  = 

Check  :  Xi  as  |.     Does  2(|)2  -  3(f)-  6  as  Of 

Does  2  •  -2^  -  V  -  5  =  0  ? 
Does  -225-  -  V-  -  5  =  0  ?    Yes. 
x2=-l.     Does  2(-  1)2  _3(-  1)-  5  =  0? 
Does  2  +  3-5  =  0?     Yes. 

Example  2.     Solve  the  equation  2^  —  3#— 3  =  0. 
Solution:     1.  a  =  2,  b=—  3,  c  =—  3. 

2.   Substituting  in  the  formula,  x  =  ~  h  ±  V6'2  ~  4  ac  : 

2  a 


-  3  ±  V9  +  24  _  3  ±  V33  _  3  ±  5.744+ 
X~  4        ~"        4        ~  4 

,.  x,  =  8-^  =  2.186+  ;  x2  =  ~2744+  =  -  .680+. 
4  4 

Check  :     2.186+  The  coefficient  of  x  is  —  f,  when  the  coeffi.- 

-  .686+  cientofx2  =  1  ;  1.5  =-(-§). 

1.500  (For  this  method  of  checking,,  s^e  §  76.) 


QUADRATIC   EQUATIONS  89 

EXERCISE   35 

Solve  the  following  equations  by  the  formula: 


1. 

a*  _  12  x  _j_  32  =  0. 

13. 

2     _    13    _    I    . 

2. 

f  +  7y-3Q  =  0. 

3 «     9  x2     18 

3. 
4. 

2z2-3z-20  =  0. 
3x2-x-A  =  0. 

14, 

2  c     11       1 
5       10     2c' 

5. 
6. 

Ay2-5y-21=0. 
20m2  + ra-2  =  0. 

15. 

x     1_  5       1 
6     3     2x     2* 

7. 

9  w2  -  13  w  4-  3  =  0. 

1  R 

6s  +  5_4s  +  4 

6. 

20m2  +  ra-2  =  0. 

7. 

9  w2  -  13  w  +  3  =  0. 

8. 

6  m2  +  m  =  3. 

9. 

4  r2  —  7  r  =  —  3. 

0. 

5a2  +  3#  =  9. 

1. 

z2      7z_4 
2       6  ~3* 

2. 

7       1_   5 
6<2     2     12  r 

4s-3       s-3 

17.    -i-  =  2*-5. 
7  — £ 

2         3      5 


18. 


iv  —  1      w     6 


19.    20^+1  _,_       1 


x  +  1  (x  +  1)2 

79.  Summary  of  Methods  of  Solving  a  Quadratic.  Four 
methods  of  solving  a  quadratic  equation  have  been  given  : 
the  graphical,  by  factoring,  by  completing  the  square,  and  by 
the  formula.  The  first  is  useful  mainly  as  a  means  of  illus- 
tration ;  the  third  is  useful  mainly  in  solving  the  general  quad- 
ratic ax2  -f-  bx  -f-  c  =  0,  and,  thus,  in  deriving  the  formula;  the 
fourth  is  used  whenever  the  solution  is  not  readily  accomplished 
by  factoring. 

Historical  Note.  Greek  mathematicians  as  early  as  Euclid  were  able 
to  solve  certain  quadratics  by  a  geometric  method,  about  which  the  student 
may  learn  when  studying  plane  geometry.  Heron  of  Alexandria,  about 
110  b.c,  proposed  a  problem  which  leads  to  a  quadratic.  His  solution  is 
not  given,  but  his  result  would  indicate  that  he  probably  solved  the  equa- 
tion by  a  rule  which  might  be  obtained  from  the  quadratic  by  completing 


90  ALGEBRA 

its  square  in  a  certain  manner.  Diophantus,  275  a.d.,  gave  many  problems 
which  lead  to  quadratic  equations.  The  rules  by  which  he  solved  his 
equations  appear  to  have  been  derived  by  completing  the  square.  He 
considered  three  separate  kinds  of  quadratics.  He  gave  only  one  root  for 
a  quadratic,  even  when  the  equation  had  two  roots. 

The  Hindu  mathematicians,  knowing  about  negative  numbers,  con- 
sidered one  general  quadratic.  Cridharra  gave  a  rule  much  like  our 
formula.  The  Hindus  knew  that  a  quadratic  has  two  roots,  but  they  usu- 
ally rejected  any  negative  roots. 

The  Arabians  went  back  to  the  practice  of  Diophantus  in  considering 
three  or  more  kinds  of  quadratics.  Mohammed  Ben  Musa,  820  a.d.,  had 
five  kinds.  He  admitted  two  roots  when  both  were  positive.  Alkarchi 
gave  a  purely  algebraic  solution  of  a  quadratic  by  completing  the  square, 
and  refers  to  this  method  as  being  a  diophantic  method. 

In  Europe,  mathematicians  followed  the  practice  of  the  Arabians,  and  by 
the  time  of  Widmann,  1489,  had  twenty-four  special  forms  of  equations. 
These  were  solved  by  rules  which  were  learned  and  used  in  a  mechanical 
manner.  Stifel,  1486-1567,  finally  brought  the  study  of  quadratics  back 
to  the  point  that  had  been  reached  by  the  Hindus  one  thousand  years 
before.  He  gave  only  three  normal  forms  for  the  quadratic  ;  he  allowed 
double  roots  when  they  were  both  positive.  Stevih,  1548-1620,  went  still 
farther.  He  gave  only  one  normal  form  for  the  general  quadratic,  as  do 
we  ;  he  solved  this  in  both  a  geometric  and  an  algebraic  manner,  giving 
the  method  of  completing  the  square.     He  allowed  negative  roots. 

EXERCISE  36 

Supplementary  Miscellaneous  Examples 

Solve  the  following  equations  by  any  of  the  preceding 
methods.  As  a  rule,  solve  by  factoring  if  possible ;  otherwise 
by  the  formula. 

1.  (3»  +  2)(2a-3)  =  (4a;-l)2-14. 

2.  y(5  y  +  22)  +  15  =  (2  y  +  5)2. 

3.    A  +  ^=_ll  5.        4  7  2 

It      3  6 

a         4  4 


7-y     y        3 


r  -  2      r  -  3      15 

3  iv         4  —  5  w 

5 

4  —  5  w         3  iv 

6 

QUADRATIC   EQUATIONS  91 


7. 

2x-l 

X 

X 

X  — 
X 

5 

10. 

-.      x  —  2  _6x 
x  +  4       5 

X  +  4: 

8. 

x-2 

ic  +  4 

7 
~3* 

11. 

a  —  5      .J  _a 
a-6            3 

x  +  5 

x-3 

a(a  - 1)      2  =  a  - 1  3*     3_l-4 

2a  +  5       3         3  4       2-*  +  3' 


a;       'x—l_x2  +  x  —  l 
16-    :t- — —  —  — 7. * 

X  —  1  X  X2  —  X 

x  x         x2  -\-2  x  —  2 


14. 


x  +  2      x  +  3      z2  +  5x  +  6 


15.  ?J1+1+  r-9  =1. 
7-r       3r-l 


16.   3«^13  =  _J t 

6  —  w       iv  —  4 


17.   1=2,  =  8+      6 


"     2v  4^-3 


18.   3w  +  g  =  1  |  2ro  +  5 
2  m  —  5  3  m  —  5 


19         5  7       =8?;2-13v-64 

2d  +  3      3v-4        6^  +  ^-12 


20.  — !—  +  — i* —  +  _15_=o. 

a-2     24(z  +  2)     4-a2 


92  ALGEBRA 

21.    Solve  the  equation  2p2x2  —  3  px  —  1  =  0. 


Solution  :  1.   Use  the  formula  x  =  ~  6±Vft   ~4ac, 

2a 
2.   a  =  2p2;  b  = -3p;  c=  -1. 


3        A  x  =  3jp  ±  V9p*  -  4  (2^)(-l)  =  3p  ±  V9 p2  +  8p'^ 
4^2  4p2 

.  ^^Sp^Viy^^SpirpVTT^B  ±Vl7 
4^2  4^2  4^ 

„  3  4-  Vl7  ,  3  -  VT7 

Hence  #i  =  — ;  and  x2  = 

4p  4p 

Solve  for  cc : 

22.  .t2  +  5  mx  -f  6  m2  =  0.  26.  .t2  4-  2  tx  =  r2  —  £2. 

23.  3a2  —  4rsc  +  5s=0.  27.  Jf/a2  =  &a  +  Z. 

24.  4  fV  +  21  te  =  18.  28.  a-2  4-  (n  -f  l)x  =  —  ft. 

25.  12a2=23^«-5e2.  29.  a;8  +  (a-  h)x -  ab  =t). 

30.  '/V.1-2  —  2(r  +  *)as  +  4  =  0. 

31.  .T2-2da;-5a;=  -  10c?. 

32.  (x  -  4)3  -  (aj  +  3)3  =  -  217. 
1  1  14 


33. 


x2  —  3  x     x2  +  4  x      15  x2 


34    2^+1      3|-_2s17 
'   3^-2^2^  +  1       4 

3^472-1      2)        1^3^  +  1      3J 
36.       1-^-1        =1+     3 


a2-4      3<>  +  2)  2-a 

37    __a a        _4t 

2it-  +  a      3  a  — 4  a      3 


QUADRATIC   EQUATIONS  93 

38.    (d2-d-2)x2-(5d-l)x  =  -6. 
__    x  —  a.x-fa     x2  —  5  a2 

Oa. 1 =  • 

x  -\-  a      a  —  x       x2  —  a2 
40.    (m  +  n)x2  +  (3 m  +  n)x  +  2m=  0. 


EXERCISE  37 

1.  Twice  the  square  of  a  certain  number  equals  the  sum  of 
15  and  the  number.     Find  the  number. 

2.  If  three  times  the  square  of  a  certain  number  be  in- 
creased by  10  times  the  number,  the  sum  is  8.  Find  the 
number. 

3.  Find  two  consecutive  numbers  whose  product  is  462. 

4.  The  sum  of  the  squares  of  three  consecutive  integers 
is  434.     Find  the  integers. 

5.  The  sum  of  a  certain  number  and  its  reciprocal  is  jf. 
Find  the  number. 

6.  Find  the  dimensions  of  a  rectangle  whose  area  is  352 
square  feet,  if  its  length  exceeds  its  width  by  6  feet. 

7.  The  denominator  of  a  certain  fraction  exceeds  twice  the 
numerator  by  2,  and  the  difference  between  the  fraction  and 
its  reciprocal  is  f-J.     Find  the  fraction. 

8.  Find  the  base  and  altitude  of  a  triangle  whose  area  is 
60  square  inches,  if  the  base  exceeds  the  altitude  by  7  inches. 

9.  Find  the  dimensions  of  a  rectangle  whose  area  equals 
that  of  a  square  of  side  18  feet,  if  the  difference  between  the 
base  and  altitude  of  the  rectangle  is  15  feet. 

10.  Find  the  dimensions  of  a  rectangle  whose  area  is  3000 
square  feet  if  the  sum  of  its  base  and  altitude  is  115  feet. 

11.  Find  the  base  and  altitude  of  a  right  triangle  if  the 
hypotenuse  is  13  feet  and  if  the  base  exceeds  the  altitude  by 
7  feet. 


94  ALGEBRA 

12.  Find  the  base  and  altitude  of  a  right  triangle  if  the 
hypotenuse  is  17  feet  and  if  the  sum  of  its  base  and  altitude 
is  23  feet. 

13.  A  fast  train  runs  8  miles  an  hour  faster  than  a  slow 
train;  it  requires  3  hours  less  for  a  trip  of  288  miles  than 
does  the  slow  train.     Find  the  rate  of  each  train. 

14.  An  automobile  party  made  a  trip  of  160  miles.  By  in- 
creasing their  average  rate  by  4  miles  an  hour,  they  can  make 
the  return  trip  in  2  hours  less  time.  Find  their  average  rate 
going. 

15.  A  crew  can  row  downstream  18  miles  and  back  again  in 
a  total  time  of  1\  hours.  The  rate  of  the  current  is  known  to 
be  one  mile  an  hour.  What  is  the  rate  of  the  crew  in  still 
water  ? 

16.  Some  boys  were  canoeing  on  a  river,  in  part  of  which 
the  rate  of  the  current  is  4  miles  an  hour  and  in  part  2  miles 
an  hour.  If,  when  going  downstream,  they  go  3  miles  where 
the  current  is  rapid  and  6  miles  where  the  current  is  slow  in  a 
total  time  of  If  hours,  what  is  their  rate  of  rowing  in  still 
water  ? 

17.  A  tank  can  be  filled  by  one  pipe  in  4  hours  less  time 
than  by  another.  If  the  pipes  are  open  together  11  hours,  the 
tank  will  be  filled.  In  how  many  hours  can  each  pipe  alone 
fill  the  tank  ? 

18.  I  have  a  lawn  which  is  60  feet  by  80  feet.  How  wide  a 
strip  must  I  cut  around  it  when  mowing  the  grass  to  have  cut 

half  of  it  ? 

Hint  :  Referring  to  the  figure,  it  is  clear  that  if 
io  =  the  number  of  feet  in  the  width  of  the  border 
cut,  then  the  dimensions  of  the  uncut  part  of  the 
lawn  are  (60  —  2  w)  and  (80  -  2  w). 

Hence,  (60  -  2  w)  (80  -  2  w)  =  £  •  60  •  80. 

Complete  the  solution. 


w 

w 

CM 
1 
O 

"~80-2w 

_^ 

L^_ 

w 

w 

w 

W 

QUADRATIC   EQUATIONS  95 

19.  A  farmer  is  plowing  a  field  whose  dimensions  are  40 
rods  and  90  rods.  How  wide  a  border  must  lie  plow  around 
the  field  in  order  to  have  completed  ^  of  his  plowing  ? 

20.  The  numerator  of  a  certain  fraction  is  2  less  than  the 
denominator.  The  reciprocal  of  the  fraction  exceeds  the  frac- 
tion itself  by  -j-jj- .     Find  the  fraction. 

21.  In  the  formula  s  =  at  +  ±gt2,  let  s  =  124,  a  =  30,  and 
g  =  32.     Find  t 

22.  From  the  formula  S  =  ^\  2  a  +  (n .—  1)  d  j ,  determine  n 
when  S  =  5,  a  =  5,  and  d  =  —  1. 

23.  The  numerator  of  a  certain  fraction  is  5  less  than  the 
denominator.  If  6  be  added  to  both  the  numerator  and  the 
denominator,  the  resulting  fraction  is  f  of  the  original  frac- 
tion.    Find  the  fraction. 

24.  A  picture  15  inches  by  20  inches  in  size  is  to  be  sur- 
rounded by  a  frame,  whose  area  shall  be  J  of  that  of  the  picture 
inclosed.     What  must  be  the  width  of  the  frame  ? 

25.  The  rate  of  one  train  exceeds  that  of  another  by  5  miles 
an  hour.  The  fast  train  makes  a  trip  of  150  miles  in  one  hour 
less  time  than  the  slow  train.     Find  the  rate  of  each  train. 

26.  A  workman  -and  his  assistant  can  do  a  piece  of  work 
together  in  3f  days.  It  would  take  the  assistant  4  days  longer 
to  do  the  work  alone  than  it  would  take  the  master  work- 
man.    How  long  would  it  take  each  alone  to  do  the  work  ? 

27.  The  area  of  a  certain  trapezoid  is  150  square  feet.  The 
upper  base  exceeds  the  altitude  ,by  2  feet  and  the  lower  base 
exceeds  the  altitude  by  8  feet.  Find  the  two  bases  and  the 
altitude  of  the  trapezoid. 

28.  Divide  30  into  two  parts  such  that  the  square  of  the 
greater  shall  equal  the  product  of  30  and  the  smaller. 

•  29.  Replace  the  number  30  of  Example  28  by  the  number  a 
and  solve  the  resulting  problem. 


.96  ALGEBRA 

IMAGINARY   ROOTS   IN   A   QUADRATIC    EQUATION 

80.   Example.     Solve  the  equation  x2  —  2x  +  5  =  Q. 

Solution  :  1.    Use  the  formula  method  of  solving  the  equation. 
a  =  1,  b  =  —  2,  c  =  5. 

2.   a; 


-  b  ±  \/P  ■ 

-  4  a< 

•  _  +  2  ±  V4  - 

4 . .  i . .  u 

2a 

2 

2±V^T6 
2 

+  2  ±  V4  - 
2 

■20 

The  question  arises,  what  does  V— 16  mean?  Is  —  4  the 
square  root  of  -16?  No,  for  (-4)2  =  +  16.  Is  4-4?  No, 
for  (+4)2  =  4-16.  Thus,  no  number  with  which  the  student 
is  acquainted  will  produce  —  16,  when  it  is  squared. 

81.  No  rational  number  raised  to  an  even  power  will  pro- 
duce a  negative  result ;  hence  an  even  root  of  a  negative  num- 
ber is  impossible  up  to  this  point.  To  avoid  this  difficulty,  a 
new  kind  of  number  is  introduced.       • 

An  Imaginary  Number  is  an  indicated  square  root  of  a  nega- 
tive number;  as,  V— 16;   V— 3;   V—  a2. 

The  numbers  previously  studied  are  called  Real  Numbers. 

82.  Every  imaginary  number  can  be  expressed  as  the  product 
of  a  real  number  and  V—  1. 

V—  1  is  indicate!  i,  and  is  called  the  Imaginary  Unit. 

Thus,  V^l6  =  V16(-  l)=        ±  4V^T  =  ±  4  i. 
V-  a"2  =  Va2(—  1)=        ±aV^l=±ai. 
V^l  =  \/5(-l)  =  ±  V5  •  V^T  =  ±i  Vft. 
Historical  Note.     The  symbol  i  for  V—  1  was  introduced  by  Euler, 
one  of  the  greatest  mathematicians  of  the  eighteenth  century. 


EXERCISE  38 

Express  the  following  in  terms  of  the  unit  i; 

1.    V-9.                      3.    V-49.r.                5 

V-25c2. 

2.    V-36.                     4.    V-100m2.              6. 

V-81a262, 

QUADRATIC    EQUATIONS  97 


7 

V-144r4. 

8. 

V-l69a4. 

9. 

V=T- 

10. 

V-ff 

11. 

V-A- 

12. 

V-y^Cl2. 

13. 

V^6. 

14. 

V-8. 

15. 

V-24. 

16. 

V-44. 

17. 

V-45c2. 

18. 

V-20a2^ 

19. 

V-50a4. 

63  y\ 


21.    Simplify  V--2^-. 


Solution:  V~T=V">^Jf ^  =  ±|- V3"  V^T  =  ±^V3- 


22.  V-|.  25.     V--\0-.  28.     V-ff' 

23.  V^-|  26.    V-  -2f.      '  29.    V— $f 

24.  V-|f  27.     V~f|.  30.     V-iH- 

83.    Addition  and  Subtraction  of  Imaginary  Numbers. 
EXERCISE   39 


1.    Add  V-4  and  V-36. 

Solution  :   V^i  +  V- 36  =  2  i  +  6  i  =  8  i. 

Note.  While  every  imaginary  number,  like  V—  4,  has  two  values,  one 
positive  and  one  negative,  in  problems  such  as  the  one  in  this  exercise, 
only  the  principal  root,  the  positive  one,  is  used,  as  in  the  case  of  surds 
(§  68). 


Simplify : 

x2  4- 

2.  V-16+V-4. 

3.  V-9+V-49. 

5/  V-100-V-64. 

6.    V-l  +V-25  -V-49. 

4.    V-81+V-25. 

'    7.    V-«2-V-4a2  -  V-9a2. 

8.    V-36 

V-  K)()x2-V-81a;2. 

9.     V-  16 

xY 

-  V  -  25  xhf  +  V  -  49  xhf. 

)8  ALGEBRA 

13.    V^20  +  V^5  -  V^45. 


14.    V-24-2V-6  +  3V-54. 


15.    V-28  +  5V-7-V-63. 
16.    Simplify  +  ?±^_?L 

SoitmoH:    1.  5±JI27=5      VE27=5      3V3.  y^ry 
2       \       4       2  2  2  2 

=5      31V3 

2  2 

_5±Siy/S 

2 

The  numbers  in  Examples  1-15  are  called  Pure  Imaginaries. 
The  sum,  or  difference,  of  a  pure  imaginary  and  a  real  number, 
§  81,  as  in  this  exercise,  is  called  a  Complex  Number. 

Simplify : 


»lWts 

■»■  i*N© 

-  £*>/# 

...  §*V3 

»■  !W^- 

-  s±vs 

84.  A  further  discussion  of  imaginary  numbers,  more 
complete,  including  a  discussion  of  the  other  fundamental 
operations  upon  imaginary  numbers,  is  given  in  Chapter 
XIV. 

85.  Meaning  of  Imaginary  Roots  of  a  Quadratic  on  the  Graph. 

Example.     Consider  the  equation  x2  +  x  -f  2  =  0. 

Solution  :  1.    Solve  the  equation  by  the  formula  : 
a  =  1,    6  =  1,    c  =  2. 


-1  ±y/TZrs=  -l  j-V_7  =  -l±i\/7 
2  2  2 


QUADRATIC   EQUATIONS 

2  2 

2.   Solve  the  equation  graphically.     (Review  rule  §  75.) 
Let"  y  =  x2  +  «  +  2  : 


99 


When  x  = 
then     y  = 

0+1 
+  2     +4 

+  2 
+  8 

+  3 
+  14 

+  2     +4 

-3 

+  8 

-4 
+  14 

3.  The  graph  has  the  same 
shape  as  the  graphs  obtained 
when  solving  other  quadratic 
equations  ;  but  the  graph  does 
not  cross  the  horizontal  axis 
at  all.  Hence,  y  or  x2  +  x  +  2 
is  never  zero  for  any  real  value 
of  x. 

This  is  characteristic 
of  the  graph  of  a  quad- 
ratic which  has  imaginary 
roots. 


\ 


+  o 


fe* 


3l 


EXERCISE   40 


Solve  the  following  equations.     Express  the  roots  in  simplest 
radical  form.     Draw  the  graphs  for  the  first  three  equations. 


1.  x2  +  x  +  l  =  0. 

2.  ic2-2z+-3  =  0. 

3.  x2-3x  +  4,  =  0 

4.  2  a2- 2  a +  1  =  0. 

5.  3m2-2m  +  2  =  0. 

6.  4c2-5c-f-2  =  0. 

7.  9r2  +  4  =  8r. 

8.  ^-££  +  1  =  0. 


10. 


m-3        2 

2  m  4- 1      m  +  7 


-  1       m  +-  1 

u.  2«  +  l+|=o. 

3        a     2 

12.  2#2-f-6da;  +  5d2  =  0. 

13.  Sx2  —  5wx  +  3w2  =  0. 

14.  5x-2-8to-5*2  =  0. 


IX.    SPECIAL  PRODUCTS  AND  FACTORING 
ADVANCED  TOPICS 

86.  In  paragraph  10  is  the  rule:  "The  product  of  the  sura 
and  the  difference  of  any  two  numbers  equals  the  difference  of 
their  squares  " ;  thus,  (x  -f  y)(x  —  y)  =  x2  —  y2  for  all  numbers  x 
and  y. 

If  z  =  2aandy  =36,  (2a+3&)(2a-3&)  =4a2-9&2. 

If  x  =  14  and  y  =  5,  (14  +  5)(14  -  5)  =  196  -  25  =  171. 

If  x  =  (a  -f  6)  and  y  =(c  +  d),  then  similarly 

[(a  +  b)  +  (c  +  d)][(a  +  6)  -  (c  +  <*)]  =  (a  +  &)2  -  (c  +  d)2. 

Likewise,  in  any  of  the  type  forms  studied  in  Chapter  II, 
the  numbers  may  be  general  number  expressions. 

Example  1.     Multiply  (a  +  b  -f-  c)  by  (a  -f  &  —  c). 
Solution  :   1.     (a  +  6  +  c) (a  +  &  -  c)  =  {(a  +  6)  +  c}{(a  +  &)  -  c} 
2.  =  (a  +  &)2  -  c2  =  a2  +  2  a&  +  b2  -  c2. 

Here  x  =  (a  +  6)  and  y  =  c. 

Example  2.     Multiply  (r  +  s  -M  —  w)  by  (r  +  s  —  t  -f-  n). 
Solution  :    1.     (r  -f  s  +  £  —  ri)  (r  -f  s  —  t  +  ft) 

2.  =  {(r  -h *)+(*-  w)}{(r  +  «)-(*-  n)}  s  (r  +  s)2  -  (I-  n)2 

3.  =  r2  +  2  rs  +  s2  -  t2  +  2  *n  -  ft2. 
Here  x  =  (r  +  s)  and  y  =  (t  —  w). 

Note.  In  such  examples,  the  rules  for  introducing  parentheses  (§  6)  are 
used.  The  various  terms  of  the  expressions  may  be  rearranged,  if  necessary, 
so  that  one  factor  becomes  the  sum  and  the  other  the  difference  of  the  same 
two  numbers,  when  the  terms  are  grouped. 

100 


SPECIAL   PRODUCTS   AJJ i> '  F&OTGRItftf  '    "    101 

EXERCISE  41 

Find  the  following  products  mentally : 

1.  S(a  +  b-)+5ll(a  +  b)-5\. 

2.  \(m  +  n)-2p\\(m  +  n)+2p\. 

3.  {10-(r  +  s)\110  +  (r  +  s)l. 

4.  \3p-(c  +  d)\\3p+(c  +  d)l. 

5.  {(c  +  2  d)  -  11  a]{(c  +  2d)  +11  ttf. 

6.  (a  —  6  4-c)(a  —  b  —  c). 

7.  (#-2/4-z)(>-2/-z). 

8.  (a2_f_a_i)(a2_a  +  i)- 

9.  (a2  +  ab  +  62)  (a2  -  ab  +  b2). 

10.  (a-f-2&-3c)(a-2&  +  3c). 

11.  (?>x  +  ±y  +  2z)(3x-4y-2z). 

12.  (a-2  4- a- 2)  (a;2 -a -2). 

13.  (a-\-r—  c  +  d)(a  +  r  +  c  —  d). 

14.  (a—  b  +  m-\-  n)(a  —  b  —  m  —  n). 

15.  (2x  +  z  —  y  +  w)(2x  —  z  —  y  —  w). 

16.  \(a  +  b)  +2(a-b)\\(a  +  b)-3(a-b)l 
Solution  :  Just  as  (re  +  2  y)  (x  —  3  y)  =  x2  —  xy  —  6  y2, 

so  {(a  +  b)  +  2(a -&)}{(«+ 6) -3(a-6)} 

=  (a  +  ft)2  -  (a  +  6)(a  -  6)  -  6(a  -  ft)2 
=  (a2  +  2  a&  +  &2)  -  (a2  -  ft2)  -6(a2-2aH  62) 
=  a2  +  2  a&  +  62  -  a2  +  &*2  -  6  a2  +  12  a&  -  6  62 
=  14  a&  -  6  a2  -  4  &2. 
Here  *  =  (a  +  6)  and  y  =  {a  -  b). 

17.  ](m4-w)-4j|(m4-M)-5S„ 

18.  \(x-y)  +  S\\(x-y)-6}. 

19.  {3aj-(y+»)H2aj-(y  +  2)(. 


102  ALGEBRA 

20.  \x  +  3y  +  15z\  \x  +  3y  -  10 z\. 

21.  \r  +  2s-3t\  jr-f  2s  +  lt\> 

22.  \3p-±(q  +  r)\  {±p- 5(q +  r)\. 

23.  \x2  +  2x  +  l\  j#2  +  2  as  -  5j. 

24.  [(a  +  6)-5]2.  26.    [2a-(c  +  d)]l 

25.  [6-f(ra-w)]2.  27.    [a  +  5  +  c]2. 

28.    From  the  result  of  Example  27,  make  a  rule  for  deter- 
mining by  inspection  the  square  of  any  polynomial. 

Find  the  following  by  the  rule  made  in  Example  28 : 

29.  [«  +  36-c]2.  31.    2r  +  s-t  +  x~]2. 

30.  [a  -b  +  c-d]2.  32.    [3«  -  b  +  2  c  -  d~]2. 

87.    General  Problems  in  Factoring. 

Example  1.     Just  as  x2  —  3a;  —  88  =  (x  —  11)  (a +  8), 

so  (a  -  2  6)2  -  3(a  -  2  6)  -  88  =  {(a  -2b)-  11}  {(a  -  2  6)  +  8} 

=  (a-2b-  ll)(a-26  +  8). 

EXERCISE  42 
Factor  completely  the  following  expressions : 

1.  («  +  &)2-c2.  7.    (x-y)2  +  2(x-y)-63. 

2.  (m-n)2-x2.  8.    (x  +  y)2  -  5(x  +  y)  -  36. 

3.  3»_(y+*)t  9.    (r  +  s)2_L.4(r_L.sy_5^ 

4.  m2-(n-p)2.  10.    (p-g)2+8(>-g)r-20r2. 

5.  (lx-2y)2-y2.  11.    (a;2-4)2-(aj  +  2)2. 

6.  (a  +  5)2  +  23(a  +  6)  4-  60.       12.    9(m  -  n)2  -  12(m-n)  -f  4. 

13.  (x  -  y)2  —  (m  —  w)2. 

14.  (a2-2a)2-f  2(a*-2a)  +  l. 

15.  (1  +  ?l2)2  -  4  W2. 

16.  (a2  +  3a)2  +  4(z2  +  3a)  +  4. 


SPECIAL  PRODUCTS  AND  FACTORING'  lW 

17.   (9  a2  +  4)2-  144  a2. 

(18.    (a2  +  7  a)2  +  20  (a2 -f  7  a) -96. 
19.    (m  +  n)2  4-  7(m  +  n)  — 144. 
20.    (a;24-»-9)2-9. 

21.  (a  +  y)3-*3.  26.    (a  +  ?/)3-f-(a-y)3. 

22.  (r  +  s)3  +  8  P.  27.    «6  -  (ar>  -h  l)3. 

23.  .(m  +  n)3 -  (m -  n)3.  28.    27 m3—(m—  n)\ 

24.  a3+(a  +  l)3.  29.    (2 a  -  bf-(a +2b)\ 

25.  a3-8(a  +  6)3.  30.   (a?  +  3  yf  -  (m  -  3  yf. 

88.    Polynomials  Reducible  to  the  Difference  of  Two  Squares. 

Certain  polynomials  may  be  put  into  the  form  of  the  differ- 
ence of  two  squares  by  grouping  certain  terms. 

Example  1.  Factor  2  mn  +  ra2  —  1  -f  n2. 

Solution  :  1.  2  mn  +  m2  —  1  +  n2  =  (m2  +  2ra«  +  n2)  —  1 

2.  =  (m  +  n)2  -  1 

3.  =  (m  +  rc  +  l)0+  w-1).     (§87) 

Example  2.  Factor  a2  -  c2  +  &2  -  d2  -  2  cd  -  2  a&. 

Solution  :  1.  a2  -  c2  +  62  -  cP  -  2  cd  -  2  a& 

2.  =(a2-2aH  62)  -  (c2  +  2 cd  +  d2) 

3.  =  (a  -  b)2  -  (c  +  d)a 

4.  =  {(a  -  6)  +  (c  +  *)}  {(a  -  6)  -  (c  +  *)} 

5.  =  (a  —  5  +  c  +  d)  (a  —  b  —  c  —  d) 

EXERCISE  43 
Factor : 

1.  a2-2ab  +  b2-c2.  6.  2mn-n2  +  l-m\ 

2.  m2  +  2mn  +  n2-p2.  7.  9a2-24a&-f-16&2-4c2. 

3.  a2-x2-2xy-y2.  8.  16  a2  -  4  ?/2 -f  20  #z  -  25  z2 

4.  a?2  —  y2  —  z2  -|-  2  yz.  9.  4  n2  -}-  m2  —  x2  —  4  mn. 

5.  &2_4  +  2a&  +  a2.  10.  4a2-66-9-&2. 


104  ALGEBRA 

11.  10  xy-9z2  +  y2  +  25  x2. 

12.  a2  -2  ab  +  b2  -  c2  +  2  cd-  d2. 

13.  a2  -  b2  -f  x2  -  y2  +  2  ax  +  2  6y. 

14.  x2  -\-m2  —  y2  —  n2  —  2  mx  —  2  ny. 

15.  2xy-a2  +  x2-2ab-b2  +  y2. 

16.  4a2  +  4a&  +  62-9c2  +  12c-4. 

17.  16?/2--36--8a?/-z2-f  «2-12z. 

18.  m2-97i2  +  25a2-62-10am4-66n. 

19.  4  a2  -  c2  -  12  ab  +  2  cd  +  9  b2  -  d2. 

20.  9  x*  -  4  x2  +  z2  -  6  a2z  -  20  a#  —  25  #2. 

89.    Trinomials  Reducible  to  the  Difference  ot  Two  Squares. 
Type  Form :  x*  +  ax2y2  +  y*. 

Example  1.     Factor  aA  +  a2b2  +  64. 

Solution  :  1.  a*  +  a2b2  +  b4  may  be  changed  into  a  perfect  square  by- 
adding  a2b2.    Adding  and  subtracting  a2b2  : 

a4  +  rt2&2  +  &4  =  (a4  +  2  «2&2  +  ft4)_  a'2&2, 

2.  .-.  a4  +  a262  +  64  =  (a2  +  62)2  -  a262 

3.  =(o2  +  62  +  «6)(a2  +  62-a6).  (§87) 

Example  2.     Factor  64  a4  -  64  a2m2  +  25  m4. 

Solution:  1.  A  perfect  square  containing  64a4  and  25 to4  is 
64  a4  —  80  a2m2  +  25  to4.  The  given  trinomial  may  be  changed  into  this 
perfect  square  by  subtracting  16  a2m2  ;  then 

64  a4  -  64  a2m2  +  25  m4  =  (64  a4  -  80  a2m2  +  25  m4)  4-  16  ahn1. 

But  this  is  the  sum  of  two  squares  and  not  factorable  in  this  form. 

2.  Another  perfect  square  containing  64  a4  and  25  m4  is  64  a4  +  80  a2m2 
+  25  to4.     Adding  and  subtracting  144  a2TO2  : 

64  a4  -  64  a2rw2  +  25  to4  =  (64  a4  +  80  a?m2  +  25  to4)  -  144  a2m2. 

3.  .-.  64  a4  -  64  a2TO2  +  25  to4  =  (8  a2  +  5  w2)2  -  ( 12  awi)2 

=  (8  a2  +  5  to2  +  12  «to)  (8  a2  +  5  to2  -  12  am). 


SPECIAL   PRODUCTS  AND  FACTORING  105 

EXERCISE  44 
Factor  the  following  trinomials  : 

1.  x*  +  x2  +  l.  13.  4  r4  -  32  r2£2  +  49  Z4. 

2.  a4  +  a2m2  +  m4.  14.  9  m4ns  —  m2n4  + 16. 

3.  c4  +  6c2  +  25.  15.  4  p*-  24  2>V  +  25r*. 

4.  y4  +  3y2  +  36.  16.  9  a4  + 17  a262  +  49  b\ 

5.  1+2^  +  9^.  17.  4#4  +  7ay  +  16.?/8. 

6.  1  -  ?-2 -f- 16  r4.  18.  9  *4  -  31  W  +  25  a,-4. 

7.  tf-lZtfif  +  ly4.  19.  16  mV  + 15  m2n2  +  25. 

8.  9m4-19m2  +  l.  20.  25  p4  +  34pV  +  49*/*. 

9.  4:y4-32y2  +  l.  21.  x4  +  4=. 

10.  25<eY-llay  +  l.  22.   ?/4  +  64. 

11.  4a44-lla2&2  +  9&4.  23.   ^  +  42/*. 

12.  9?>i4  f  14m2w2  +  25n4.        24.   4^  +  1. 

90.    Certain  polynomials  can  be  factored  by  grouping  their 
terms. 

Type  Form :  ab  +  ac+ bd+cd=  (a  +  d)(b  -f  c). 

Example  1.     Just  as  ax  -\-  bx=(a  +  b)x,  (§  5) 

so  a(x  +  y)+  b(x  +  y)  =  (a  +  b)(x  +  y).  (§  5) 

Example  2.     Factor  6  a?  — 15  a2  —  8  x  +  20. 
Solution:  1.    6^-15z2-8x+20  =  (6£8-15x2)-(8a;-20)         (§6) 
2.  =3z2(2z-5)-4(2a;-5)(§  15,  a) 

S.  =  (3x2-4)(2x-5). 


106  ALGEBRA 

EXERCISE  45 
Factor : 

1.  2a(x  +  y)-3(x  +  y).  11.  2  +  3a-  8a2-  12a3. 

2.  5m(r  +  s)  +  2w(r+s).  12.  3 Xs  +  6 x2  +  x -f  2. 

3.  3i?(2a-?/)  —  r(2x-y).  13.  10ma -15ftz- 2m +  3?i. 

4.  8(£  +  w)  —  m(£  +  w).  14.  a3x  +  a&ca;  —  a2by  —  b2cy. 

5.  a(6  +  c)  -  cZ(6  +  c).  15.  a2bc  -  ac?d  +  ab2d  -  bed2. 

6.  a&  +  an  +  6m  +  mw.  16.  30  a3  —  12  a2  —  55  a  +  22. 

7.  ax-ay-\-bx-by.  17.  56  -  32 a  +  21^-12^. 

8.  ac  —  ad  —  be  +  &d.  18.  3  aa;  —  a?/  +  9  bx  —  3  &y. 

9.  a3  +  a2  +  a+l.  19.  4  ars  +  x2y2  -  4  y3  -  16  xy. 
10.   4^—5^  —  4^  +  5.  20.  rt  —  rn  —  sn  +  st. 

21.  ar  +  as  +  6r  +  6s  —  cr  —  cs. 

22.  ax  +  ay  —  az  +  bx  —  6z  +  6y. 

23.  am  —  6m  —  cp  +  ap  —  cm  —  6p. 

24.  or*  —  xz2  +  #2y  +  xy2  +  ys  —  yz2. 

25.  2  ax  +  ca  +  3  by  —  2  ay  —  3  bx  —  cy. 

91.    The  Sum  or  the  Difference  of  Two  Like  Powers. 

Type  Form :  an  ±  b\ 

By  actual  division,  as  in  §  9,  c : 

(a*  -  b*)  +  (a  -  b)  =  a3  +  a26  +  ab2  +  68. 
(a4-  bA)  +  (a  +  b)=as-a2b  +  a&2-  63. 
(a5-  65)-*- (a  -  6)=  a4  +  a36  +  a2b2  +  a&3  +  64. 
(a6  +  66)-s-(a  +  6)=  a4  -  a36  +  orb2  -  a&3  +  64. 


SPECIAL  PRODUCTS  AND  FACTORING  107 

The  following  rule  may  be  verified  in  the  same  manner:  * 
Rule.  —  I.    Letting  n  represent  any  positive  integer: 

1.  an  —  bn  is  always  exactly  divisible  by  a  —  b. 

2.  an  —  6"  is  exactly  divisible  by  a  +-  b  when  n  is  even. 

3.  an  +-  6*  is  never  exactly  divisible  by  a  —  b. 

4.  an  +-  bn  is  exactly  divisible  by  a  +  b  when  n  is  odd. 

II.   In  the  quotient : 

1.  The  signs  are  all  plus  when  a  —bis  the  divisor. 

2.  The  signs  are  alternately  plus  and  minus  when  a  +-  b  is  the 
divisor. 

3.  The  exponent  of  a  in  the  first  term  is  1  less  than  its  expo- 
nent in  the  dividend,  and  decreases  by  1  in  each  succeeding  term 
until  it  becomes  1. 

4.  The  exponent  of  &  is  1  in  the  second  term,  and  increases  by  1 
in  each  succeeding  term  until  it  becomes  1  less  than  its  exponent 
in  the  dividend. 

Example  1.     Divide  a7  —  b7  by  a  —  b. 

Solution  :  1.   By  I,  1,  a7  —  67  is  exactly  divisible  by  a  —  6. 

2.  By  II,  1,  3,  and  4, 

qT""&7  =  «6  +  a66  +  a4&2  +  a368  +  a2&4  +  a&5  _f_  &6. 

a  —  b 

Example  2.    Factor  32  a?  +  243. 
Solution:  1.   32  x*>  +  243  =(2  x)5  +  35. 

2.  This  expression  is  of  the  type  an  +  6n,  where  a  =  2  sc,  6  =  3,  and 
w  =  5.     By  I,  4  and  II,  2,  3,  and  4, 

32a:5  +  243=(2a;  +  3)[(2a;)*-(2a:)3.3+(2x)2.32-(2a;).88  +  34] 

=  (2  x  +  3)  (16  tf  -  24  Xs  +  36  re2  -  54  x  +  81). 
Check  :  Let  x  =  1.     Then  32  a5  +  243  =  32  +  243  =  275  ;  also, 
(2  x  +  3)  (16  &  -  24  xs  +  36  sc2  -  54  sc+81)  =  (2+3)  (16-24+36-54+81) 

=  5  •  65  =  275. 

*  These  rules  are  proved  in  §  257. 


108  ALGEBRA 

The  method  of  factoring  binomials  of  the  form  an  ±  bn  given 
in  this  paragraph  must  be  used  frequently.  However,  if  the 
prime  factors  (§  14)  of  a  binomial  of  this  form  are  desired, 
proceed  as  in : 

Example  3.     x6  -  y*  =  (as8  f  f)  (ar3  -  f) 

=  (x  +  y)(tf  ~xy  +  y2)(x  -  y)(x?  +  xy  +  y2). 

Note  1.    By  §91,  xt  —  y*  =(x  —  y)  (x5  +  x*y -\ f-?/5). 

The  second  factor  is  not  prime  however. 

Note  2.  Whenever  the  binomial  is  the  difference  of  two  even  powers,  it 
may  be  treated  as  the  difference  of  two  squares. 

Example  4.     x9  +  y9  =  (st?f  +  (iff  =  (or3  -f  tf)(x«  -  a?f  +  f) 

=  (x  +  y)(x*  -®y  +  y2K^  -  ^f  +  y6)- 

Note  1.    By  §  91,  x?  +  ?/9  =  (z  +  ?/)  (a;8  -  a;?*,  +  x6y2  +  ...  +  ?/8). 
The  second  factor  is  not  prime  however. 

EXERCISE  46 
Find  the  following  quotients : 
t,    ^Z..  6.    21^.  11.    1-1(5«4 

7.  *±£  12. 

x  +  z3 

8.  i  +  i5.  13. 

1  +  a 

9.  |H£  14 

2  — a 

10    mT~<  15 

a&  -f-  c  m  —  71  3  c  +  2  d 

Factor  the  following  expressions,  if  possible  : 

16.  27x?  —  8y3  18.    a,-5-?/10.  20.    res«  —  y*. 

17.  a4-2/4.  19.   32  -m5.  21.    a7-f&r. 


a4- 

-64 

a  +  b 

a4- 

-b* 

a- 

■b 

ar5- 

y\ 

a>  — 

■y 

m5- 

-i 

m  - 

-l 

a666 

-c6 

l  +  2a 

81a?4-?/4 

3a?  +  # 

16  -xA 

2-x 

a*  -  243  ar5 

a  — 3  a? 

81c4-16d4 

Xs  +     x2  - 

-2\x-2 

xs  -  2  x2 

3x* 

Zx*- 

-6x 

6x-    2 

6z-12 

SPECIAL   PRODUCTS   AND   FACTORING  109 

22.  32  +  r\  25.   xw-y10.  28.   64  -  a6. 

23.  l  +  32m6.  26.    x8  +  y\  29.   a8 -256. 

24.  z6  +  ?/6.  27.    m5-243.  30.   32  a5  +  243  b5. 

SUPPLEMENTARY  TOPICS 

92.    The  Remainder  Theorem  makes  it  possible  to  find  the 
remainder  in  certain  division  problems  by  a  short  process. 
It  is  known  that  dividend  =  divisor  x  quotient  -f-  remai?ider. 
Suppose  that  x*  -J-  x2  —  2  is  divided  by  x  —  2 ;  then  : 

xs  +  X2__2=  (x  -2)  •  Q  +  B.  x2  +  3  x  +  6 

Let  x  =  2  ;  then  : 
8+4-2=0.  Q  +  B. 

.-.  10  =  0  •  Q  +  B. 

.:  10  =  i?. 
That  is,  the  remainder  is  10. 
Check  :  See  solution  on  right. 

10  =  B. 

At  the  left,  the  correct  remainder  was  obtained  by  substitut- 
ing 2  for  x  in  the  given  expression.     This  suggests  the 

Remainder  Theorem.  If  a  rational  and  integral  polynomial 
(§  12)  involving  x  be  divided  by  x—  a,  the  remainder  may  be 
found  by  substituting  a  for  x  in  the  given  polynomial. 

Proof  :  1.     The  polynomial  (containing  x)  =  (x  —  a)  •  Q  4-  B. 

2.  .-.  The  polynomial  (x  replaced  by  a)  =  (a  —  a)  ■  Q  +  B. 

3.  ,\  The  polynomial  (x  replaced  by  a)=  0  •  Q  +  B  =  B. 

Example  1.  Find  the  remainder  when  xi  —  3  x  +  5  is  di- 
vided by  x  —  3. 

Solution:    1.    Comparing  x  —  a  and  x  —  3,  a  must  be  3. 
2.    Substituting  3  for  x  in  x*  -  3  x  +  5, 

i2  =  34-3.3  +  5=  81  -9  +  5  =  77. 

Example  2.     Find  the  remainder  when  the  divisor  is  a; +  3. 
Solution  :    1.    Comparing  x  —  a  and  x  +  3,  a  must  be  —  3. 
2.    Substituting  —  3  for  x  in  x*  —  3  x  +  5, 

22=z(-3)4-3-  (_3)+5  =  81 +9  +  5  =  95. 


110  ALGEBRA 

EXERCISE  47 
Find  the  remainders  when : 

1.  x5  +  2  x4  —  7  is  divided  by  x  —  1 ;  by  x  + 1. 

2.  2#3+3arJ  —  4  #  -f  5  is  divided  by  a;  —  5  ;  by  x  -f  5. 

3.  a4 4-2^-6  is  divided  by  x  +  2;  by  x  —  2. 

4.  m3  —  2  m2  —  4  is  divided  by  m  —  3 ;  by  ra  4-  4. 

5.  #4  —  y3  4-  y2  —  y  +  6  is  divided  by  ?/  -f-  3 ;  by  y  —  2. 

93.  Synthetic  Division  is  a  short  process  for  finding  the 
quotient  as  well  as  the  remainder  when  a  polynomial  containing 
x  is  divided  by  a  binomial  of  the  form  x  —  a. 

Consider  the  two  solutions : 


Solution  (a):  Solution  (6): 

5 a3 -11  a2-    6x- 
5      +15      +12+18 


5s2  +    4s  4  6 aj  +  3|5aj3-lla;2-    6  x  -  10 

«  -  3  5  x8-  11  x2-  6  x  -10 


5"^rf  fl,  5  4  4+  6||+  8. 
4 z2  -  12  g  Quotient :  5a;2  +  4a;  +  6. 
;r-  __  «  q                Remainder :    4  8. 

6x-18 
4    8 
The  method  of  performing  the  solution  (6)  : 

(1)  —  3  of  the  original  divisor  is  changed  to  4  3. 

(2)  5  Xs  -f-  x  =  5  x2.     Place  5  in  the  third  line. 

(3)  4  3.4  5=415.  Add  the  product,  4  15,  to  —11.  Place  the 
sum,  4  4,  in  the  third  line. 

(4)  +  3  •  4  4  =  4  12.  Add  the  product,  4  12,  to  -  6.  Place  the  sum, 
4  6,  in  the  third  line. 

(5)  4  3.46=418.  Add  the  product,  4  18,  to  -  10.  Place  the 
sum,  4  8,  in  the  third  line. 

(6)  The  numbers  5,  4  4,  and  4  6  are  the  coefficients  of  the  quotient. 
Since  5  sc8  -4-  x  =  5  z2,  the  full  quotient  is  5  x2  4  4  x  +  6.  The  last  number 
of  the  third  line,  +  8,  is  the  remainder. 


SPECIAL  PRODUCTS   AND  FACTORING  111 

A  partial  explanation  follows  : 

(1)  In  step  (1),  —  3  is  changed  to  +  3.  This  permits  addition  in  steps 
(3),  (4),  and  (5)  instead  of  the  customary  subtraction.  Thus,  in  solution 
(a),  when  —  15  x2  is  subtracted  from  —  11  x2,  the  result  is  4  x2 ;  in  solu- 
tion (6),  when  +  15  ic2  is  added  to  —  11a;2,  the  result  is  again  4  a;2. 

(2)  In  step  (3),  +4  below  the  line  represents,  first,  the  coefficient  of 
the  first  term  of  the  remainder  as  in  solution  (a).  When  4x2  is  divided 
by  x  the  quotient  is  +  4  x,  so  that  4  may  properly  be  considered  also  the 
coefficient  of  the  second  term  of  the  quotient.  Similarly  in  the  case  of 
+  6  in  step  (4). 

Example  2.     Divide  7  xA  -  29  M  -  3  tA  by  x  +  2 1. 
Solution.     Change  x  +  2t  to  x  —  2t. 


x-2t 


7  x4  +    0  tx3  -  29  t2x2  +  0  fix  -  3  tt 
7      -ut     -f28«2     +2*3-4*4 


7      _14*     _       t2     +2«3||-7*4 

Quotient  :  7  xz  —  14  tx2  —  t2x  +  2  t3.        Remainder :     —  7  tf4. 

Note  1.  "When  powers  of  x  are  missing,  supply  them  with  coefficients  zero, 
as  in  this  example. 

EXERCISE  48 
Divide  by  synthetic  division : 

1.  x?-2x2  +  2x-5byx-l. 

2.  2x?-4:X2  +  6x-15hy  x  +  1. 

3.  f-Sy  +  10bjy-2. 

4.  z4  +  5!?  +  15z2-25byz  +  2. 

5.  ^-32  by  t-2. 

6.  3m4-25m2-18by  ra-3. 

7.  4a3  +  18a2  +  50  by  a  +  5. 

8.  6c4  +  15^  +  28c  +  5by  c+3. 

9.  3  x3  +  4  ma;2  —  2  m2a;  —  5  m3  by  a;  —  m. 
10.   4  a4  -  15  5V- 4  64  by  x  +  2b. 

Note.  In  §§  92  and  93,  two  short  processes  for  finding  the  remainder  in 
certain  division  problems  are  given.  Each  is  important.  The  second  has  the 
further  advantage  of  determining  the  quotient  as  well. 


112  ALGEBRA 

<H    The  Factor  Theorem  makes  it  possible  to  factor  certain 
polynomials. 

Example  1.     Is  x  —  1  a  factor  of  x3  +  x2  —  2  ? 
Solution  :   1.   If  x  -  1  is  a  factor,  the  remainder  when  x?  +  *2  —  2  is 
divided  by  x  —  1  will  be  zero. 

2.  By  the  remainder  theorem,  B  =  1  +  1  —  2  ~  0.     (a  =  1.) 

3.  .\  x  —  1  is  a  factor  of  x3  +  x2  —  2. 

This  example  illustrates  the 

Factor  Theorem:  If  a  rational  and  integral  polynomial  (§  12) 
involving  x  becomes  zero  when  x  is  replaced  by  a,  then  the  poly- 
nomial has  x  —  a  as  a  factor. 

Proof  :  Considering  the  given  polynomial  the  dividend  and  x  —  a  the 
divisor,  the  remainder  will  be  the  value  of  the  dividend  when  x  is  replaced 
by  a.  According  to  the  conditions,  this  value  is  zero  ;  hence  the  remain- 
der will  be  zero  and  the  division  exact.  Therefore  x  —  a  is  a  factor  of  the 
polynomial. 

In  applying  the  factor  theorem: 

1.  Determine  mentally,  if  possible,  some  number  a  for  which 
the  given  polynomial  becomes  zero.     (Factor  Theorem.) 

2.  Divide  the  polynomial  by  x  —  a  by  synthetic  division 
(§  93).  This  division  will  give  further  assurance  that  the  re- 
mainder is  zero,  and  will  also  determine  the  other  factor,  the 
quotient.     This  factor  may  often  be  factored. 

Example  2.     Find  the  factors  of  x3  +  3  x2  —  4  x  —  12. 
Solution:    1.  When x  —  1,  then  (mentally)  sc3-f  3a;2  —  4 x—  12=  — 12. 
,\  x—\  is  not  a  factor  of  the  polynomial. 

2.  When  x  =  —  1,  xs  +  3  x2  -  4  x  —  12  =  —  6.    .-.  *  +  1  is  not  a  factor. 

3.  When  x  =  2,  x3  +  3  x2  -  4  x  -  12  =  -  0.     .-.  x  -  2  is  a  factor. 

4.  Dividing  by         x  +  2 1  x*  +  3  x2  -    4  x  -  12     Remainder  =  0. 
synthetic  division  :  11+2      +10     4-12     .-.  x  -  2  is  a  factor. 

1+5      +    6 1|        0    The   other   factor   is 
z2  +  5x  +  6. 

5.  .-.  Xs  +  3  x2  -  4  x  -  12  =  (x  -  2)  (x2  +  5  x  -  0) 

=  (x-2)(x  +  2)(x  +  3). 


SPECIAL   PRODUCTS   AND   FACTORING  113 

EXERCISE  49 
Factor  by  the  factor  theorem : 

1.  aJ2  +  a._6  6t  2y*  +  y2-3. 

2.  x?-2x2-x  +  2.  7.  tf-2z2  +  3. 

3.  x*  +  x2  —  4  a  —  4.  8.  r3  +  4r!  +  6r  +  4. 

4.  a3 -6  a2 +  11  a -6.  9.  £4  +  *3  -  2  £2  -  t  +  1. 

5.  x3  —  x2  —  9  x-\-9.  10.  m4-5m3  +  5m2-)-5m-6. 

11.   ar*  +  mx2  —  2  m3. 

Let  x  =  m  ;  m:i  +  m8  —  2  m3  =  0. 

.*.  x  -  m  is  a  factor. 

Find  the  other  factor  by  division. 

12.  3x*+p2x-±pz.  14.   a;8  +  3  fa?*  —  4 *». 

13.  ar*  —  5  ra;2  +  6  r3.  15.   xA  —  ex3  —  7  cV  +  (Ab  +  6  c4. 

EXERCISE  50 

Miscellaneous  Examples 

In  the  following  list  of  examples,  the  types  of  factoring 
studied  in  this  chapter  will  be  used.  Before  taking  up  the  list 
of  examples,  review,  if  necessary,  the  rules  for  obtaining  the 
H.C.F.  and  the  L.C.M.  of  two  or  more  expressions  in  Chapter 
II  and  for  operations  with  fractions  in  Chapter  III.  The  ex- 
amples marked  with  an  asterisk  (*)  depend  upon  the  supple- 
mentary topics  in  §  92  to  §  94  and  should  be  omitted  if  these 
paragraphs  are  not  studied. 

Factor  the  following  expressions  : 

1.  a2bc  +  ac2d  -  ab2d  -  bed2.  6.  15  ac +18  ad -35  be -A2bd. 

2.  a2-(b  +  e)2.  7*  x*  +  ±x2  +  x-6. 

3.  4  abb2  +  4  a2bb.  8.  3aGb2-3ab7. 

4.  x*  +  x2y  +  xy2  +  y3.  9.  a4 -22  a2 +  81. 

5.  1  —  a8.  10.*  x*  —  x2-±x  —  ±. 


114  ALGEBRA 

11.  (??-5xf  -2(^-5^-24  21.   128 -m7. 

12.  16aJ*-76ajy+81y*.  22.   2a26c-2  63c-46V-2  6c8. 

13.  a8-2a3+l.  23.   a;10  +  2  or5  +  1. 

14.  9(m+w)2-12(m+n)  +  4.  24.  a3  -  5  a2-  10  -t-  2 a. 

15.  32z5  +  ?/10.  25*  a4-4a3-f  a2-6a+8. 
16.*   a3-a2-5a-3.  26.   a10-l. 

17.  9a2+252/2-16z2+30a^.  27.  a6-a4  +  a2-l. 

18.  aW  +  aY-bW-tff.  28.  a7  +  a4  -  a8  - 1. 

19.  m4-625.  29.  (x2-y2-z2)2-±y2z\ 

20.  a6 -7  a3 -8.  30.  (a?+b*)-2  ab(a  +  V). 

Find  the  H.  C.  F.  and  the  L.  C.  M.  of  the  following : 

31.  3a8-21a2-a  +  7,anda2  +  6a-91. 

32.  ac  +  ad  —  be  —  bd,  and  a2  —  6  ab  +  5  62. 

33.  a2  +  b2  -  c2  +  2  a&,  and  a2  -  b2-  c2  +  2  be. 

34.  m3  —  4  m,  m3  +  9  m2  —  22  m,  2  m4  —  4  m3  —  3  m2  +  6  m. 

35.  3a8-a2H3a6-62,  27  a3-  63,  9a2-6ab  +  62. 

36.  16  m4  —  n4,  16  m4  —  8  m2n2  +  n4,  2  ma;  -f-  2  m?/  —  wa;  —  ny. 

37.  a3  —  a2a?  —  ax2  +  a^,  3  a3  —  3  a2a;  -f-  5  ax2  —  5x*, 
38  *   a^  +  *  ~  2,  and  x2  +  a?  —  2. 

39.*   a^  —  a2  —  a;  —  2,  and  3^  +  ^—6. 

40.   a^  +  Saa^-a^-Sa3,  and  a?  +  2  aar5 -  a2a; - 2  a8. 

Simplify  the  following  fractional  expressions : 

4       x2  —  y2 -\-  z2  -f  2  a%  4„     ax—bx  —  ay  j- 6y 

4m3-10m2-6m+15  ..    2  ac  -  2  fa?  -  aa*  +  bd 

'  6ma+8r^-9m-12'  d2-4c2 


SPECIAL   PRODUCTS   AND   FACTORING  115 

45. 


46. 
47. 

48. 


a-2  +  ff2-z2-f  2ay  §  ag  -  y2  +  2  yg  -  z2 

x2jty2_z2_2Xy    '    Xl-y*-2yZ-Z2' 

o»  +  M    A,  c^  +  c2 


('-.- 


a_6  +  c^        a2_«5+^ 

a?-b2-c2-2bc  :  a  -  &  -  c 
a2-&-c?  +  2bc'  a  +  b  —  c 

ar  —  as  4-  br  —  bs    ar  +  as  +  br  +  bs 


r2  _  s2  a2  +  2  a&  +  62 

49.    Solve  the  equation  :  a2  +  aa;  —  3  a6  —  3  6a;  =  0. 

Solution  :   1.     x2  +  ax  —  3  a&  —  3  bx  =  0. 

2.  Factoring  :   a;(»  +  a)  —  3  b(x  +  a)  =  0. 

(x  -  3  &)(«  +  a)  =  0. 

3.  /.  x  =  3  6,  or  x  =  -  a.  (§  74) 

Solve  the  following  equations : 

50.  x2  +  ax  —  ab  —  6x  =  0.       52.    aar*  —  bx  —  acx  +  6c  =  0. 

51.  x2-2mn-  ra2-n2  =  0.     53.   aaj*-2  da;-6d  +  3aa;  =  0. 

54.  3  cma^  +  3  mnx  —  ap#  —  mp  =  0. 

55.  acfcc2  -f  cdx  +  aea;  -f  ce  =  0. 

95.  An  equation  of  the  first  degree,  having  one  unknown, 
has  one  root ;  an  equation  of  the  second  degree  has  two  roots. 
In  general,  an  equation  of  the  nth.  degree,  having  one  un- 
known, has  n  roots. 

The  roots  of  equations  of  degree  higher  than  the  second  are 
not  obtained  readily,  except  in  particular  equations  which  may 
be  solved  partially  at  least  by  the  factoring  method. 

Example  1.     Find  the  roots  of  x4  - 13  x2  -j-  36  =  0. 
Solution  :  1.     Factoring,  (x2  —  4)  (a;2  -  9)  =  0. 
2.    .-.  x2  -  4  =  0  ;     .-.  x2  =  4  ;     .-.  x  =  2,  or  x  =  -  2. 
Also,    a;2 -9  =  0;    .-.  z2  =  9  ;    .-.  x  =  3,  or  x  =-3. 


116  ALGEBRA 

Example  2.     Solve  the  equation  if  +  2  y2  —  4  y  + 1  =  0. 
Solution:  1.   Factoring  by  the  factor  theorem, 

0/-i)Q/*  +  3*,-i)  =  a 

2.  .•.  ?/  —  1  =  0,  or  y  =  1. 

Also,         ^  +  8»-M0;-.,.dliVEi=-8±ii, 

Z  'A 

...  y  =  - 3  ±3.605.  y  =  302+5  or  _  3302+. 
z 
Hence,  */i  =  1  ;  y2  =  .302+  ;  y3  =-  3.302+. 

EXERCISE  51 
Solve  the  equations : 

1.  a?4 -26  a2 +  25  =  0.  9*   m3  -  19  m-30  =  0. 

2.  m4- 11m2 +18  =  0.  10>   z3 -  z2 -Sz  +  2  =  0. 

3.  ^-15?y2-16  =  0.  11.*   ^_5f-2  =  0. 

4.  4*4-17*2  +  4  =  0.  12.   ^-1=0. 

5.  9^  +  14^-8  =  0.  13.   ^-8  =  0. 

6*   ^-7^  +  6  =  0.  14.    r3-5?-2  =  5-r. 

7.   aj8+2a?-9«-18  =  0.  15.    xb-xA  -  16  a?  +  16  =  0. 

8.*   f-lSy-  12  =  0.  16.*  x4-x3-5»2-.c-6  =  0. 

Solve  for  a; : 

17.  x4-m2x2-n2x2+m2n2=0.         19*   2  a3-  3a?x  +  a8  =  0. 

18.  a^  +  &»2-ac2z-6c2=0.  20.   a4-r4  =  0- 

Remark.    The  graphical  solution  of  equations  of  higher  degree  is  consid- 
ered in  §  2(37. 


X.      QUADRATIC    EQUATIONS    HAVING    TWO 
VARIABLES 

GRAPHICAL  SOLUTION 
96.   Graph  of  a  Single  Equation. 

Example  1.     Draw  the  graph  of  y  —  a?  —  0. 
Solution:   1.   Solve  the  equation  for  y  :    y  =  x2. 
2. 


When  x  s= 

0 

1 

o 

:; 

4 

5 

-1 

-2 

-3 

-4 

-5 

then  y  = 

0 

1 

4 

9 

1(3 

25 

+  1 

+  4 

+  « 

+  16 

+  25 

3.  This  curve  is  a  Parabola. 

4.  The  coordinates  of  any 
point  on  the  parabola  satisfy 
the  equation.  The  coordinates 
of  A  are:  x  =  _45 

and  y  -  20+. 

Substituting  in  y  =  x2  :  does 
20+  =(- 4.5)2? 

Does  20+  =  20.25  ?  Yes, 
approximately.  The  coordi- 
nates should  satisfy  the  equa- 
tion of  the  graph.  Since  the 
graph  cannot  be  absolutely  ac- 
curate, and  since  the  coordi- 
nates of  a  point  on  the  graph 
cannot  be  read  exactly  from 
the  graph,  the  coordinates  de- 
termined may  not  exactly  sat- 
isfy the  equation. 


¥            =F 

5C              :c 

T          "  t 

t 

1 

V 

t 

i 

_r 

:  -JL        "  40 

zL  _ 

7 

t 

t 

3 

t 

jt 

-C           5 

j 

v          i£ 

7 

i      : 

t 

— V 

y 

::    5        5 

:       5       : 

.___    \      _£ 

±  _    _ 

:  V  :    : 

::::?    ::  : 

i 

-                    ^            5- 

\ 

r 

.XL_           ~^-'- 

cj^-1              --x 

-5  -4    -3    -2     - 

2      3      4      5 

*-- 

117 


118  ALGEBRA 

Example  2.     Draw  the  graph  of  x2  +  y2  =  25. 


Solution  :  1.   Solve  the  equation  for  y  :  y  =  ±  V25  —  x2. 
2. 


When  x  = 

0 

+  1 

+  2 

+  3 

+  4 

+  5 

y  = 

±  V25 

±5 

±V24 
±4.8 

±V21 

±4.5 

±Vlo 

±4 

±V9 
±3 

±V0 
0 

3.    When  x  is  negative,  y  has  the  values  given  by  the  corresponding 
positive  values  of  x.    Thus,  when  x  is  —  3,  y  is 

±  V25-(-3)2  =  ±  V25-9  =  ±  VlG  =  ±  4. 

Notice  that  for  each  value  of  x,  y  has  two  values  ;  thus,  when  x  is  +  4, 
y  is  either  +  3  or  —  3.    Hence,  both  (4,  3)  and  (4,  —  3)  are  on  the 

graph. 

For  x  greater  than  5,  y  is 
imaginary  ;  thus,  when  a: =6, 


-V 

X'                                                 _Q_                                          X 
.-.5  _|~4    =3    -2    -                       2       3      4        5" 

y  =  ±  V25  —  36  =  ±  V- 11. 

This  means  that  there  are 
not  any  points  on  the  graph 
for  values  of  x  greater  than  5. 
The  square  roots  required 
in  step  2  may  be  obtained 
either  by  the  method  of 
§  64,  or  from  the  table  of 
square  roots  constructed  in 
§65. 

4.  This  curve  is  a  Circle. 
Every  equation  of  the  form 
x2  +  y2  =  r2,  is  a  circle  with  its  center  at  the  origin  and  its  radius  equal 
to  r. 

Example  3.     Draw  the  graph  of  9  x-  +  25  f  =  225. 

225  -  9  x2 


Solution  :  1 .  y'2  = 


25 


y=  ±$V226-0X*. 


2.   When     x  =  2,    y  =  ±  \  V225  -  36  =  ±  \  V189  =  ±  $(3  V21)  = 
±K3x4.5)  =  ±K13.5)  =  ±2.7. 


QUADRATIC  EQUATIONS  HAVING  TWO  VARIABLES    119 


When  x  = 

0 

+  1 

+  2 

+3 

+  4 

+  5 

+  6 

then  y  = 

\/227 
5 

±¥ 

±3 

±^V216 
±2.9 

±iVl89 

±K13.5) 

±2.7 

±^Vl44 

±¥ 

±2.4 

±1.8 

Vo 

0 
0 

±^V^99 

imag'y 
imag'y 

For  negative  values  of  x,  y  has  the  values  given  by  the  corresponding 
positive  values  of  x.     (See  Example  2,  step  3.) 

Notice  that  for  each  value  of  x,  there  are  two  values  of  y.     ' 


|) 

— 

►-«, 

*' 

/ 

/ 

/ 

1 

y 

1 

I 

s 

_ 

i 

-3 

-o 

- 

2 

il 

<: 

1 

k 

^ 

> 

\ 

V 

/ 

\ 

v 

"V 

V 

S 

■«~ 

-3 

' 

3.  This  curve  is  an  Ellipse.  Every  equation  of  the  form  az2  +  by2  =  c, 
where  a,  b.  and  c  are  positive,  and  a  not  equal  to  b,  has  for  its  graph  an 
ellipse. 

Example  4.     Draw  the  graph  of  9  a2  —  4  #2  =  36. 
9z2-36 


Solution  :   1.  y*  as 


;  .•.w=±|Vz2-4. 


2.   When  x  =  l,  y  =  ±|Vl  —  4=±f  V— 3  ;  .-.  y  is  imaginary. 


When  x  = 

0 

+  1 

+  2 

+3 

+4 

45 

+  0 

then  y  = 

±|V=4 
imag'y 

±|V"^3 
imag'v 

±|Vo 

0 

±fV5 
±3  3 

±|V12 
±5.1 

±|V21 
±6.8 

±|V32 
±8.4 

120 


ALGEBRA 


For  negative  values  of  x,  y  has  the  values  given  by  the  corresponding 
positive  values  of  x.     (See  Example  2.) 


Y 

s, 

s 

_ 

s 

\ 

^ 

A 

L 

_- 

-  6 

_ 

-t 

t 

-4 

- 

') 

A 

\ 

f 

-j 

\ 

\ 

j 

/ 

I 

i 

3.   This  curve  is  a  Hyperbola.    Every  equation  of  the  form  ax2  —  by2  =  c, 
where  a,  6,  and  c  are  positive  numbers,  is  a  hyperbola. 


EXERCISE  52 

Draw  the  graphs  for  the  following  equations;    name  the 
curves  obtained. 


1.  a*  +  y*  =  36. 

2.  y  =  Sx2. 

3.  x2  =  6y. 

4.  o^  +  4?/2  =  36. 


5.  x2-4,y2  =  36. 

6.  xy  =  4. 

7.  x2  +  tf  =  55.  ' 

8.  4z2  +  2/2  =  16. 


97-    Solution  of  a  Pair  of  Simultaneous  Quadratic  Equations. 


Example  1.     Solve  the  pair  of  equations 


a»+y*=25.     (1) 
x-y+l  =  0.  (2) 


Solution  :  1.    The  graph  of  equation  (1)  was  drawn  in  Example  2  of 
§96;  it  is  the  circle  of  radius  5,  with  center  at  the  origin. 


QUADRATIC  EQUATIONS  HAVINTG  TWO  VARIABLES  121 

2.  The  graph  of  equation  (2)  is  found  as  in  §  46.  It  is  a  straight 
line  (§  4(5).     When  as  =  0,  y  =, I  j  when  x  =  2,  y  =  3. 

3.  Since  points  A  and  B  are  on  both  graphs,  their  coordinates  should 
satisfy  both  equations. 

A  —  (3,  4)  ;  B  =(—  4,  —  3).  When  the  coordinates  of  ^4  and  of  B  are 
substituted  in  the  equations,  it  is  found  that  they  satisfy  the  equations. 

.\  x  =  3,  y  =  4,  and  x  =  —  4,  y  =  —  3  are  solutions  of  the  pair  of 
equations. 


Nun 

--«5     -4^3    i^2L                    2,3456 

it                 IL^ L 

Note.  Since  the  graph  of  every  linear  equation  having  two  variables  is  a 
straight  line  (§  47),  and  since,  as  the  student's  subsequent  courses  in  mathe- 
matics will  show,  the  graph  of  every  quadratic  equation  having  two  variables 
must  be  one  of  the  curves  discussed  in  §  9b',  it  is  clear  that,  as  a  r.ule,  a 
quadratic  and  a  linear  equation  with  two  variables  will  have  two  common 
solutions,  for  a  straight  line  will,  in  general,  meet  such  curves  in  two  points. 

The  straight  line  might  touch  the  curve  at  only  one  point,  thus  giving  only 
one  solution ;  or  it  might  not  touch  the  curve  at  all,  thus  not  giving  any  real 
solution. 

Example  2.     Solve  the  pair  of  equations  :  I    ,     «   2  )o\ 

Solution  :  1.  The  graph  of  equation  (1)  is  the  circle  of  radius  5 
(see  Ex.  2,  §  96).  The  graph  of  equation  (2)  is  the  ellipse  of  the 
figure. 

2.    The  points  of  intersection  of  the  graphs  are  : 

A:  (4,3);  B:  (-4,3);  C:  (-4,  -3);D:  (4,  -3). 


122 


ALGEBRA 


3.   Substituting  these  values  of  x  and  y  in  equations  (1)  and  (2),  it  be- 
comes clear  that  the  equations  have  four  common  solutions. 


. 

*% 

/'^•""*"        n 

-Ss 

jtr      I 

^s 

rf  B 

A1 

^Sr 

VX       " 

It 

SSit" 

1  > 

i                        -'- 

X      -                                                            Q 

X 

-R    -7  -fi  -5     -4    -3   r2    "I               1 

9      3      < 

5     6 

7 

f 

..i. 

/ 

Vl                                 ' 

J  V         ~ 

\A                             -2 

IZZ 

^ 

^4      : 

N^c                 ... 

1 

? 

^v 

*<£ 

££-^.-4       „ 

--*£ 

^s. ■-* 

S 

^ - 

-5 

7 

Note.  Two  quadratic  equations,  having  two  variables,  will  have  four 
common  solutions,  in  general.  This  becomes  clear  when  the  graphs  of  §  96, 
which  result  from  such  equations,  are  combined  in  pairs.    For  example: 


However,  there  are  other  possibilities.  Thus,  the  ellipse  might  intersect 
only  one  branch  of  the  hyperbola  in  such  manner  as  to  give  only  two  real 
solutions ;  or  it  might  not  intersect  it  at  all,  giving  no  real  solutions. 


2. 


QUADRATIC  EQUATIONS  HAVING  TWO  VARIABLES    123 

EXERCISE  53 

Solve  the  following  pairs  of  equations  graphically : 

faj»4-y»=100.  5     (x2  +  y2  =  50. 

\x-y  +  2  =  0.'  [xy=-7. 

±x2  +  y2  =  61.  t3x2  +  4y2  =  76. 

2x-y  =  l.  '    l32/2-lla;2  =  4. 

r^_?/=_9.  7  |^-2/2  =  16. 

'     \2x-y  =  3.       I  '    \y2  +  x  =  ±. 

4      f«+f--«L  |4^  +  2/2  =  36. 

*     ljcy=-7.  '     \x2-y2  =  -16. 

9.   Draw  the  graph  of  — -  +  ^-  =  1.     On  the  same  sheet,  draw 

the  three  graphs  obtained  from  the  equation  x2  =  y  +  7c,  when 
k  is  made  successively  6,  2,  and  —  6. 

Remark.  The  four  curves  studied  in  this  chapter,  the  circle,  the  parab- 
ola, the  ellipse,  and  the  hyperbola  are  called  conic  sections,  for  each  may  be 
derived  by  intersecting  a  circular  cone  of  two  nappes  by  a  plane.  A  special 
study  of  these  curves  is  made  in  a  later  course  in  mathematics,  analytic 
geometry. 


XI.     SIMULTANEOUS   EQUATIONS 
INVOLVING  QUADRATICS 

98.  A  set  of  equations  having  two  or  more  variables  are 
called  Simultaneous  Equations  if  each  equation  is  satisfied  by 
the  same  set,  or  sets,  of  values  of  the  variables. 

99.  A  set  of  equations  that  are  solved  as  simultaneous 
equations  will  be  called  a  system  of  equations. 

100.  A  pair  of  simultaneous  linear  equations  (§  47)  having 
two  variables  have  one  common  solution  (§  51).  The  com- 
mon solution  is  readily  obtained  by  the  addition  or  subtrac- 
tion method  (§  53),  or  by  the  substitution  method  (§  54)  *of 
elimination. 

101.  Pairs  of  simultaneous  equations  occur  of  which  one 
or  both  are  of  degree  higher  than  the  first. 

Thus,  in  (a)  below,  equation  (1)  is  of  the  first  degree  and  (2)  is  of 
the  second  ;  in  (6),  equation  (1)  is  of  the  second  degree  and  (2)  is  of  the 
third. 

,  ,     f3z+4*,  =  5.         (1)  j^-2^6.       (1) 

(a)    \2x^-3xy  =  7.     (2)  W    \x*  +  yS=5.         (2) 

Many  such  combinations,  even  with  two  variables,  are  possi- 
ble. Only  in  special  cases,  however,  are  the  common  solutions 
readily  obtained.     A  few  such  cases  will  be  considered. 

102.  Case  I.     One  Linear  and  One  Quadratic  Equation. 

ra.2  +  2/2_|_6a-16  =  0.      (1) 
Example.     Solve  the  system  :    ]i_i_2a; i/  =  0  (2) 

Solution  :  1.    From  (2),  V  =  2  x  +  *•     (8) 

2.    Substituting  in  (1),  z2+  (2a;+  l)2  +  6*  -  16  =0.  (4) 

124 


SIMULTANEOUS   EQUATIONS 


125 


(5) 
(6) 


3.  Simplifying  (4),  x2  +  2x-3  =  0. 

4.  Solving  for  x,  x  =  1,  or  x  =  —  3. 

5.  Substitute  in  (2) :  when  x  =  1,  1  +  2  —  y  =  0  ;  .-.  y  =  3. 

when  £=—  3,  1  —  6  —  ?/  =  0 ;  .-.  y  =  —  5. 
The  solutions  are :  x  =  1,  ?/  =  3  ;  £  =  —  3,  y  =  —  5. 
The  solutions  may  be  checked  by  substitution. 


Note.  One  linear  and  one  quadratic  equation  having  two  variables  have, 
in  general,  two  common  solutions.  The  graphical  solution  of  a  particular 
pair  of  equations  of  this  type  is  given  in  Ex.  1,  §  i>7. 


5. 


8. 


9. 


m2  +  mn 
m—  n 


t2  =  — 19. 


-7. 

x  +  y  =  -3. 
54. 


f*+Jf 

\xy  = 


6a 


\x*-xy  +  y2 
\x-y  =  -3. 

r  0^  +  2/2  =  101. 
\x  +  y  =  -9. 

<x*  +  xy  +  y2  =  39. 

\x~+y  =  -2. 


t2y  +  2x  = 
\2x+2y= 


2  y  +  2x  =  5  xy. 
5. 


10. 


11. 


12. 


13. 


14. 


15. 


16. 


EXERCISE  54 

Solve  the  following  systems  of  equations : 

(a?+b2=113. 

'    |a-6  =  -l. 

<5x*-3f  =  -7. 
'    \y  +  2x  =  7. 

\2x?-±xy  +  3tf^VL 
\x-3y  =  5. 


3c  +  2d  =  —  2. 
cd  +  8c  =  4. 

7a2  +  10a&  =  -8. 
8. 


j  V  a'  +  1U  aO 
15a+4&  = 

r 


jr-1. 

icy  =  a2  -f  a. 

a^  +  2/2  =  2(a2-f  62). 
x  -+-  y  =  2  a. 


3+3  =  4 
a     6     5 


a  +  6  =  16. 


^  +  -  =  1. 


r     *  =10 
t      r      3' 
3r-2t  =  -12. 


126  ALGEBRA 


HOMOGENEOUS  EQUATIONS 

103.  An  equation  is  a  Rational  Equation  if  the  variable  does 
not  appear  under  a  radical  sign. 

104.  A  rational  and  integral  (§  11)  equation  is  Homogeneous 
if  all  of  its  terms  are  of  the  same  degree  (§  44)  with  respect 
to  the  variables. 

Thus  :  x2  —  3  xy  +  y2  =  0  is  a  homogeneous  equation  ;  x2  —  xy  +  y2  =  5 
is  homogeneous  except  for  the  constant  term ;  x2  —  3  y  =  2  y2  is  not 
homogeneous. 

105.  Case  II.  Quadratic  Equations  Homogeneous  Except  for 
the  Constant  Term. 

Example!.     Solve  the  system :    J    0  +    ^  ~~      *  ^J 

[2C?  +  xy  +  2y2  =  lo.      (2) 

Solution  :  1.  Eliminate  the  constant  terms : 

M4(l):*  4x2  +  12y2  =  \12.  (3) 

M7(2):  7x2  +  1xy+Uy2  =  112.  (4) 

(4)  -  (3)  :  3  x2  +  7  xy  +  2  y2  =  0.  (5) 

2.    Solve  (5)  for  x  in  terms  of  y : 

(3»  +  y)(a:  +  2y)=0;  .\  x  =-  |,  or  i  =-  2y. 

o 

Substitute  —  J  for*  In  (1)  i      . •.  ^  +  3  y2  =  28. 

3  J  9 


2/2  +  27  y2  =  9  •  28  ;  28  y*  =  9  .  28  ;  y2  =  9  ;  y  =  ±  3. 
3    ,     3 


When  y  =  3  :     aj  =  —  ^  —  —  -  =  —  1.     .  •.  #  =  —  1,  y  =  3  is  a  solution. 


When 


y  =  —  3  :  z  =  —  ^  =  —  (  —  j=b  1,  .-.  x  =  1,  y  ——  3  is  asolution. 

3.    Substitute  -  2  y  for  a;  in  (1).     .-.  4  y2  +  3  y2  =  28. 

.-.  7y2  =  28;  y2  =  4;  y  =  ±  2. 
When  y  =  2  :      se=  —  2?/=—  2  •  2  =  —  4.   .•.  se=  —  4,  y=2  is  a  solution. 
When  y  =—2:  £  =  — 2  y=  — 2  •  —  2=4.    .-.  sc=4,  ?/  =  —  2  is  a  solution. 

*See§53forM4(l). 


\2x2-xy=2.  7      \x>  +  5xy-y2  =  -7. 


SIMULTANEOUS   EQUATIONS  127 

Check  :  These  four  solutions  are  readily  checked  by  substitution  in 
equations  (1)  and  (2). 

Note  1.  In  case  one  equation  does  not  have  a  constant  term,  solve  it  im- 
mediately for  one  variable  in  terms  of  the  other  as  the  equation  (5)  in  step  2. 

Note  2.  A  system  consisting  of  two  quadratic  equations  has,  in  general, 
four  solutions. 

Note  3.  The  graphical  solution  of  a  particular  pair  of  equations  of  this 
type  is  given  in  Ex.  2,  §  97. 

EXERCISE  55 

Solve  the  following  systems : 

(3cd  +  2d2  =  -7.  <a2  +  ab  +  b2=G3. 

[c2-2cd  =  30.  '     \a2-b2  =  -27. 

±x2  +  y2  =  l0.  \x2  +  3xy-2y2  =  -4:. 

[n2  +  3mn  =  2.  \2x2-xy  =  2S. 

'     l9m2+27i2  =  9.  '     \x2  +  2y2  =  l%. 

lr*  +  rh  =  75.  I2x2-3xy  +  5y2  =  38. 

'     U*  +  »*  =  125.  \3x2  +  xy-10y2  =  Q. 

|  x2  -f-  xy  =  —  6.  f  m2  —  2  mn  =  84. 

I  xy  —  y2  =  —  35.  1 2  mn  —  n2  —  —  64. 

106.  Equivalent  Systems.  One  system  of  equations  is  equiv- 
alent to  another  when  the  common  solutions  of  each,  system 
are  the  solutions  of  the  other  system. 

107.  Case  III.     Systems  Reducible  by  Division.     A    given 

system  may  sometimes  be  reduced  by  division  to  an  equivalent 

system  in  which  the  equations  are  of  lower  degree. 

i  x>  *—  %m  —  K(5 

Example.     Solve  the  system :    I    „ 

J  \  &  +  Xy  +  y2  =  28.  (2) 

Solution:   1.    Dividing  (1)  by  (2):  X  —  y  =  2.  (3) 


(i) 


2.    Form  the  new  system  : 


x2  +  xy  +  y*  =  28.  (2) 

x  -  y  =  2.  (3) 

3.   Solve  the  new  system  by  the  methods  of  Case  I : 
z  =  4,  y  =  2  ;  and  x  =—  2  ;  y  =—  4. 


128  ALGEBRA 

Check  :  These  two  solutions  are  readily  checked  by  substitution  in  the 
equations  (1)  and  (2). 

Note  1.  Whenever  possible,  divide  one  equation  of  the  given  system  by 
the  other,  member  by  member,  and  form  a  new  system  consisting  of  the  quo- 
tient equation  and  the  divisor  equation. 

Note  2.  The  full  theory  underlying  this  type  of  example  belongs  in  a  more 
advanced  text  and  is  therefore  omitted. 

108.  Number  of  Solutions.  In  Case  I  (§  102)  two  solutions 
and  in  Case  II  (§  105)  four  solutions  are  generally  obtained. 
The  following  rule  for  determining  the  number  of  solutions  of 
any  system  of  equations  having  two  variables  is  given  without 
proof : 

Rule. — Two  integral  equations,  having  two  variables,  whose 
degrees  are  m  and  n  respectively,  have ,  in  general  mn  common 
solutions. 

Thus,  a  cubic  (third  degree)  equation  and  a  quadratic  equation  would 
have  six  common  solutions.  If,  however,  the  system  could  be  reduced  to 
a  simpler  system,  as  in  the  example  of  §  107,  then  the  number  of  solutions 
would  be  determined  by  the  degrees  of  the  equations  forming  the  new 
system. 

EXERCISE  56 
Solve  the  following  systems  of  equations : 

\x2-y2  =  56.  „     \a3  +  b3  =  -335. 


1  x  +  y  m  14.  I  a2  -  ab  +  b2  =  67. 

fa;4-?/ =  240.  „     fm3-w3  =  -117. 


2'    \x2  +  y2  =  20.  "    lm-w  =  -3. 

lx>-f  =  133.  jTSe  +  tf-i 

*'    \x-y  =  7.  '    I  27  <!■  +  #  =  98. 

1^-2/3  =  37.  9     1^  +  ^  =  9^. 

\m*  +  Ty  +  tf-m.  \x  +  y  =  6. 

^  +  2/3  =  _217.  ia    1^  +  ^  =  504. 

x  +  y  =  -7.  '    lie2—  xy  +  tf  =  84. 


SIMULTANEOUS  EQUATIONS 


129 


11. 


12. 


13. 


\x-y  =  S. 

I  a?y  -  xy2  =  30. 

1  x  -  3  y  =  6. 

f  a2  -  a#  +  3  x  =  8. 
\xy-y2+3y  =  4;. 


14. 


15. 


ar5  +  f  =  26  a3 
a;  +  ?/  =  2  a. 

i  +  i  =  91. 

r  ? 

1/     0 


109.  Miscellaneous  Types  and  Methods.  Many  systems  of 
equations  which  cannot  be  solved  by  the  methods  already 
given  may  be  solved  by  combining  the  equations  so  as  to  ob- 
tain a  linear  equation  or  an  equation  of  the  form  a??/  =  a 
constant. 

Example  1.     Solve  the  system:  f^^2 *+2 ^=23.    (1) 

J  1  xy=6.      (2) 

Solution  :  1.     M2  (2) :  2  xy  =  12.  (3) 

2.   Adding  (1)  and  (3)  :  x2  +  %xy  +  y2  +  2x  +  2y  =  35.  (4) 

.'.  (x  +  y)2  +  2(x  +  y)-S5  =  0. 
.'.  («+y  +  7)(aj  +  y-5)=0.     (§87) 
.-.  x  +  y  =  -  7,  orx  +  y  =  5.      (§74) 


(5) 


3.   Form  the  systems  :  A  :    \  X  +  V  ~  a  7'     B:    \X  + 
[       xy  =  6.  \       i 


y  =  5. 
a-y  =  6. 


1. 


4.    Solving  ^4  :  »  =  —  1,  y  =  —  6  ;  or  x  =  —  6,  y  = 
Solving  JB ;  as  =  3,  y  =  2  ;  orx  =  2,y  =  3. 

Check  :   The  four  solutions  check  when  substituted  in  equations  (1) 
and  (2). 

(7)1j     I    TThTL  —I—  7l   —  7  (\  \ 

m+n=5+mn.    (2) 

Solution:   1.   Square  (2)  :  m2  +  2 row  +  w2  =  25  +  10 row  +  ro2w2.  (3) 

2.    Subtract  (1)  from  (3)  :  mn  =  18  +  10  mn  +  ro2w2.  (4) 

.-.  ro2w2  +  9row  +  18  =  0.  (5) 
.-.  (ron  +6)(mw  +  3)=0;  .-.  row  =—  6,  or  row  =—3.    (§  74) 

ro  +  w  =  5  +  row.     „ 
row  =  —  6. 


3.   Form  the  systems  :  A : 


'•     t>  .  J  ro  +  w  =  5  + 1 
row  =  —  3. 


130 


ALGEBRA 


4.    Solving  A  :  m  =  2,  n  =  —  3  ;  or  m  =  - 
Solving  B :  m  —  3,  n  =  —  1 ;  or  m  =  - 
Check  :  The  four  solutions  check  when 
and  (2). 

EXERCISE  57 

Solve  the  following  systems : 
xy  =  12. 


-  3,  n  =  2. 

-  1,  n  =  3. 
substituted  in  equations  (1) 


2. 


6. 


8. 


10. 


11. 


12. 


y2  =  40. 

A2B2+2$AB-m=0. 
2A  +  B  =  11. 

2w2-5t2  =  13. 
15t2  +  w2  =  2L 

m2  +p2  =  1. 

4x2  +  y2  =  6l. 
2ar2  +  32/2  =  93. 

4v2  —  5va?=19. 

^^+0^  =  6. 

3^_5^  +  2i2=-3. 


fSr  —  on 

\r-t  =  l. 

(c?  +  cd  +  d2 
ic-d  =  19. 


97. 


x*  +  tf  =  756. 
tx2  —  xy  +  y'2  =  63. 

>2-s2  =  3. 
[j»  =  -  2. 

V  +  262  =  47  +  2a. 
^a2_2b2=  -7. 


xy  =  a2—  1. 
2  a. 


7?, 


13. 


14. 


15. 


16. 


17. 


18. 


19. 


20. 


21. 


22. 


=  10 
n     m-\-n      3 


'm_±_n.m 


m 

m2  -f  n2  =  45. 

V-2/3  =  3a2  +  3a  +  l. 
x~y  =  l. 

'7v2-5vt-3t2=36. 
>v2  +  3vt  +  t2=-4. 

15 


x  -f  y     2  x 


V 


x  —  y     x-\-2y      4     * 
x-3y«-2. 

i1+m)=m- 

K1+sfo)=48a 

f  m  —  v  =  —  31. 

[  mi?  =  — 150. 

fa^  +  /=:7a8. 

[x  +  y  =  a. 

{x2  +  y2=2a2-2ab  +  b\ 
{{2x2-y2=a2  +  2ab-b2. 

x2  +  y2  =  5(a2  +  b*). 
4^--y2=5a(3a-46). 

|8ic2-ll2/2  =  8. 
il2ic24-13  2/2  =  248. 


f  18 


23. 


r  —  s     r  + 
r-2s  =  l. 


SIMULTANEOUS  EQUATIONS 
30 


131 


14 


=  8. 


x2-y2  =  16. 
y2  —  14  =  x. 

xi-\-x2y2-{-y4  =  91. 
,x2  +  xy  +  y2  =  l3. 
Hint  :  See  §  207. 

fo4  +  aa62+64  =  133. 

U2- 


24. 


25. 


26 


27. 


a&  +  62  =  7. 


y      x         % 

x  +  y  =  l. 


Hint  :  Clear  of  fractions ;  divide 
(1)  by  (2). 


28. 


m2  —  mn  =  27  n. 
mn  —n2  =  3m. 
Hint  :  M3  (2)  ;  add  ;  factor. 

[y2  =  3y-x. 
Hint:  Find  (l)-(2). 


y(x  —  a)  =  2  a&. 
#(#  —  6)  =  2  ab. 


fx2y  —  x=  — 14. 
Uy  +  a2  =  148. 

5m2-9?i2=-121. 
7n2-3m2  =  105. 

f  mn  —  (m  —  n)  =  1. 
lm27i2  +  (w-n)2==13. 

a2-ab-12b2  =  S. 
a2  +  ab  -10  62  =  20. 

2tf*-Sa#=  -4. 

4  ##  —  5  ?/2  =  3. 

|i2  +  5^-^=-7. 
\t2  +  3tw-2w2=-4;. 

\xy  +  (x-y)  =  —5. 
\xy(x  —  y)  =  —  84. 

,9x2-xy  —  y  =  51. 
-5xy  +  y2  +  3x  =  $l. 


31. 


32. 


33. 


34. 


35. 


36. 


37. 


EXERCISE  58 

1.  Find  two  numbers  whose   sum  is  15  and  the  sum  of 
whose  squares  is  113. 

2.  Find  two  numbers  whose  difference  is  9  and  the  sum  of 
whose  squares  is  221. 

3.  Find  two  numbers  whose  difference  is  7  and  whose  sum 
multiplied  by  the  greater  gives  400. 

4.  The  difference  of  the  squares  of  two  numbers  is  16  and 
the  product  of  the  numbers  is  15.     Find  the  numbers. 


132  ALGEBRA 

5.  The  sura  of  the  squares  of  two  numbers  is  52;  the 
difference  of  the  numbers  is  one  fifth  of  their  sum.  Find 
the  numbers. 

6.  The  difference  of  the  cubes  of  two  numbers  is  218 ;  the 
sum  of  the  squares  of  the  numbers  increased  by  the  product 
of  the  numbers  is  109.     Find  the  numbers. 

7.  If  the  product  of  two  numbers  be  multiplied  by  their 
sum,  the  result  is  —  6 ;  and  the  sum  of  the  cubes  of  the  num- 
bers is  19.     Find  the  numbers. 

8.  Find  two  numbers  whose  difference  is  4  and  the  sum  of 
whose  reciprocals  is  f . 

9.  The  sum  of  the  terms  of  a  fraction  is  13.  If  the  numera- 
tor be  decreased  by  2,  and  the  denominator  be  increased  by  2, 
the  product  of  the  resulting  fraction  and  the  original  fraction 
is  ^-.     Find  the  fraction. 

10.  Find  the  number  of  two  digits  in  which  the  units'  digit 
exceeds  the  tens'  digit  by  2,  and  such  that  the  product  of  the 
number  and  its  tens'  digit  is  105.     (See  §  172,  First  Yea)- Algebra.) 

11.  The  sum  of  the  squares  of  the  two  digits  of  a  number 
is  58.  If  36  be  subtracted  from  the  number,  the  digits  of  the 
remainder  are  the  digits  of  the  original  number  in  reverse 
order.     Find  the  number. 

12.  Find  the  number  of  two  digits  such  that,  if  the  digits 
be  reversed,  the  difference  of  the  resulting  number  and  the 
original  number  is  9,  and  their  product  is  736. 

13.  The  area  of  a  rectangular  field  is  216  square  rods,  and 
its  perimeter  is  60  rods.  Find  the  length  and  width  of  the 
field. 

14.  The  hypotenuse  (§  73)  of  a  certain  right  triangle  is 
10  feet,  and  the  area  of  the  triangle  is  24  square  feet.  Find 
the  base  and  altitude  of  the  triangle. 


SIMULTANEOUS   EQUATIONS  133 

15.  Find  the  dimensions  of  a  rectangle  whose  diagonal  is 
2V10  inches  and  whose  area  is  12  square  inches. 

16.  A  rectangular  field  contains  2\  acres.  If  its  length 
were  decreased  by  10  rods,  and  its  width  by  2  rods,  its  area 
would  be  less  by  one  acre.  Find  its  length  and  width.  (See 
p.  145,  First  Year  Algebra.) 

17.  The  altitude  of  a  certain  rectangle  is  2  feet  more  than 
the  side  of  a  certain  square ;  the  perimeter  of  the  rectangle  is 
7  times  the  side  of  the  square,  and  the  area  of  the  rectangle 
exceeds  twice  the  area  of  the  square  by  32  square  feet.  Find 
the  side  of  the  square  and  the  base  of  the  rectangle. 

18.  If  the  length  of  a  rectangular  field  be  increased  by  2 
rods  and  its  width  be  diminished  by  5  rods,  its  area  becomes 
24  square  rods ;  if  its  length  be  diminished  by  4  rods  and  its 
width  be  increased  by  3  rods,  its  area  becomes  60  square  rods. 
Find  its  length  and  width. 

19.  A  man  has  two  square  lots  of  unequal  size,  together  con- 
taining 74  square  rods.  If  the  lots  were  side  by  side,  it  would 
require  38  rods  of  fence  to  surround  them  in  a  single  inclosure 
of  six  sides.     Find  the  length  of  the  side  of  each. 

20.  A  and  B  working  together  can  do  a  piece  of  work  in 
6  days.  It  takes  B  5  days  more  than  A  to  do  the  work.  Find 
the  number  of  days  it  will  take  each  to  do  the  work  alone. 

21.  Find  the  sides  of  a  parallelogram  if  the  perimeter  is 
24  inches  and  the  sum  of  the  squares  of  the  number  of  inches 
in  the  long  and  short  sides  is  80. 

22.  One  of  two  angles  exceeds  the  other  by  5°.  If  the  num- 
ber of  degrees  in  each  is  multiplied  by  the  number  in  its 
supplement,  the  product  obtained  from  the  larger  of  the  given 
angles  exceeds  the  other  product  by  the  square  of  the  number 
of  degrees  in  the  smaller  of  the  given  angles.    Find  the  angles 


134  ALGEBRA 

23.  Two  angles  are  supplementary.  The  square  of  the  num- 
ber of  degrees  in  the  larger  angle  exceeds  by  4400  the  prod- 
uct of  the  number  of  degrees  in  one  angle  by  the  number  in 
the  other  angle.     Find  the  number  of  degrees  in  each  angle. 

24.  The  difference  in  the  rates  of  a  passenger  train  and  a 
freight  train  is  10  miles  per  hour.  The  passenger  train  re- 
quires 1  hour  more  for  a  trip  of  175  miles  than  the  freight 
train  requires  for  a  trip  of  100  miles.     Find  the  rate  of  each. 

25.  A  crew  can  row  upstream  18  miles  in  4  hours  more  time 
than  it  takes  them  to  return.  If  they  row  at  two  thirds  of 
their  usual  rate,  their  rate  upstream  would  be  1  mile  an  hour. 
Find  their  rate  in  still  water,  and  the  rate  of  the  stream. 

26.  The  area  of  one  square  field  exceeds  that  of  another 
square  field  by  1008  square  yards  ;  the  perimeter  of  the  greater 
exceeds  one  half  of  that  of  the  less  by  120  yards.  Find  the 
side  of  each  field. 

27.  The  tens'  digit  of  a  certain  number  exceeds  the  units' 
digit  by  1.  If  the  square  of  the  given  number  be  added  to 
the  square  of  the  number  with  the  given  digits  in  reverse  order, 
the  sum  is  585.     Find  the  number. 

28.  If  the  digits  of  a  number  of  two  figures  be  reversed,  the 
quotient  of  this  number  by  the  given  number  is  If,  and  their 
product  is  1008.     Find  the  number. 


XII.     THE   THEORY   OF   QUADRATIC   EQUATIONS 

110.    The  Sum  and  the  Product  of  the  Roots. 

The  general  quadratic  equation  is : 

aa?+bx  +  c  =  0.  (1) 

Divide  both  members  by  a : 

^  +  ^.a-f£  =  0.  (2) 

a  a 

The  roots  of  (1)  are : 


6+V6»-4oe  5-vV-4«e.  ( 

2a  2a 

ri  +  r2  =  — —  = (4) 

2a  a 

(_  6)2 _  ( Vfc2  -  4  ac)2  _  62 - (b2  -  4  oc)  _  4  oc      c       ... 

?i  •  r2  = — — — —  - — -  —  -  •     {o) 

4o2  4  a2  4  a2      a 

Rule.  —  In  the  general  quadratic  equation  ax1  -f  bx  +  c=  o : 

1.  The  sum  of  the  roots  is •  From  (4). 

a 

2.  The  product  of  the  roots  is  -•  From  (5). 

a 

3.  If  the  coefficient  of  x2  is  made  1,  the  coefficient  of  x  is  the 
negative  of  the  sum  of  the  roots,  and  the  constant  term  is  the 
product  of  the  roots.  From  (2),  (4),  (5). 

Example  1.  Find  the  sum  and  the  product  of  the  roots  of 
the  equation  2a?2— 9#  —  5  =  0. 

Solution:    1.    a  =  2;  b=—  9;  c  =  —  5. 

2.  ...ri  +  „=_»=_^2  =  +  |i  ™,=5  =  ZlS. 

a  2  2  a        2 

Note.  The  first  part  of  this  rule  justifies  the  method  of  checking  solutions 
of  quadratic  equations  recommended  in  §  7(5. 

135 


136  ALGEBRA 

EXERCISE  59 

Find  by  inspection  the  sum  and  the  product  of  the  roots; 
check  examples  1,  2,  3,  and  7  by  finding  the  roots : 

1.  a?  +  7a  +  6-=0.  5.  9r-21r2  +  7  =  0. 

2.  ra2-ra  +  12  =  0.  6.  4- ?/- 6  2/2  =  0. 

3.  3c2-c-6  =  0.  7.  2  a2 +  3^-5^  =  0. 

4.  12  2/2-42/  +  3  =  0.  8.  14  a2  +  8  tx  +  21  *2  =  0. 

9.    One  root  of  4^  —  #  —  5  =  0  is   —  1.     Find  the  other 
root. 

Solution  :   1.   r±  =—  1.     Let  r2  be  the  second  root. 
2.   n  +  r2  =  +  J  ;  /.  -  1  +  r2  =  i,  or  r2  =  1^  =  £. 

Check :   Does  n-  r2= ?  i.e.  does  —  1  •  -  = ?    Yes. 

4  4        4 

10.  One  root  of  3  x2  +  7  x  —  6  =  0  is  f .     Find  the  other. 

11.  One  root  of  7  q2  +  20  g  +  12  =  0  is  -  2.     Find  the  other. 

12.  One  root  of  15  ra2  +  28  ra  =  32  is  f.     Find  the  other. 

13.  One  root  of  3  x2  -  2  kx  =  33  k2  is  -  3  &.     Find  the  other. 

14.  One  root  of  4j92  —  15  xp  —  4  x2  =  0  is   —  4j?.     Find  the 
other. 

15.  Find  k  so  that  one  root  of  x2  —  5  x  -f  A;  =  0  may  be  7. 
Solution  :   1.    r\  +  r2  ==  5 ;  .*.  r2  +  7  =  5,  or  r2  =  —  2. 

2.    *  =  ri  •  r2  ;  .-.  A;  =  7  •  -  2  =  -  14. 

Check  :   If  x2  —  5  x  -  14  =  0,  then  (at  -  7)  (x  +  2)  =  0.      .-.  as  =  7,  or 
-2. 

16.  Find  &  so  that  one  root  oi2x2  —  Sx—k  =  0  may  be  3. 

17.  Find  k  so  that  one  root  of  3a2  —  7  x  —  2k  =  0  may  be 
-2. 

18.  Find  w  so  that  one  root  ofa52-|-7a;-f4ft  =  0  may  be  5. 

19.  Find  p  so  that  the  roots  ofor2-f-3a;-f-j?  =  0  shall  be  equal. 


THE   THEORY  OF  QUADRATIC   EQUATIONS       137 

20.  Find  r  so  that  the  roots  of  3x2  —  5x  +  r  =  0  shall  be 
equal. 

111.  Formation  of  Equations  Having  Given  Roots.  There  are 
two  methods  of  forming  a  quadratic  equation  which  shall  have 
given  roots. 

Example  1.     Form  the  equation  whose  roots  shall  be  i  and 


-f. 


Solution  :   1.  Let  the  coefficient  of  x2  be  1 ;  then  by  §  228  the  equation 
x2  —  (ri  +  r2)  x  +  rir2  =  0. 

3.   .-.  the  equation  is  : 

*2-(-i)*+(-|)  =  0'  orz*  +  |-|  =  0.     (§228) 

Multiplying  both  members  by  8, 

8  x2  +  2  x  -  3  =  0. 
Check  :  The  given  roots,  if  substituted,  will  satisfy  the  equation. 

Example  2.     Form  the  equation  whose  roots  shall  be  —  9 
and  2. 

Solution  :    1.   If  x  as  -  9,  then  x  +  9  =  0  ;  K  x  =  2,  then  x  -  2  =  0. 

2.  .♦.  (x  +  9)(x-2)  =  0,  or  x2  +  7x-  18=0. 

It  is  clear  that  this  equation  has  the  given  roots. 

Note.    This  second  method  may  be  used  also  to  form  an  equation  having 
three  or  more  roots. 

EXERCISE  60 
Form  the  equations  whose  roots  shall  be : 

1.  2,3.  4.   12,-5.  7.   if. 

2.  -3,-6.  5.   2,f.  8.   3m,  -5m. 

3.  6,  -  9.  6.   1,  - 1  9.   4 1,  - 1 1. 

10.  -fc,  fc.  13.   2a-b,  2a  +  b. 

11.  2,  3,  -  5.  14.   3  +  V5,  3  -  V5. 
*2.   a  +  3m,  a-3m.                    15.   2  +  3V2,  2-3V2. 


138  ALGEBRA 

DETERMINATION  OF  THE  CHARACTER  OF  THE  ROOTS 

112.  Classification  of  Numbers.     The  numbers  considered  in 
this  text  to  this  point  are : 

(A)  Eeal  numbers. 

1.  Kational  Numbers:    (a)  integers  (positive  and  negative); 
(b)  fractions  whose  terms  are  integers. 

2.  Irrational  numbers :  (a)  quadratic  surds  (§  67)  ;  (6)  surd 
expressions,  such  as  2  +  V3. 

(B)  Imaginary   Numbers :    (a)   pure    imaginaries    (§    83) ; 
(6)  complex  numbers  (§  83). 

113.  It  is  often  necessary  to  determine  the  character  of  the 
roots  of  a  quadratic. 

Thus  the  roots  of  2  x2  -  8  x  +  3  =  0  are  4=bv^0> 

Since  10  is  positive,  the  roots  are  real  numbers. 
Since  10  is  not  a  perfect  square,  the  roots  are  irrational. 
Since  VlO  is  added  in  one  root  and  subtracted  in  the  other,  the  roots 
are  unequal. 

Hence  the  roots  are  real,  irrational,  and  unequal. 

It  is  possible  to  determine  the  character  of  the  roots  how- 
ever without  determining  the  roots  themselves. 

For  the  general  quadratic  ax2  -+-  bx  -f  c  =  0,  the  roots  are 


_-&  +  V52_.4ac  -&-V62-4ac 

• 

Rule  1.  —  If  ft2  —  4  ac  is  positive,  the  roots  are  real  and  unequal. 
They  are  rational  if  fc2  —  4  ac  is  a  perfect  square,  and  irrational  if 
ft2  —  4  ac  is  not  a  perfect  square. 

2.  If  ft2  —  4  ac  equals  zero,  the  roots  are  real  and  equal. 

3.  If  ft2  —  4  ac  is  less  than  zero,  the  roots  are  imaginary. 
ft2  —  4  ac  is  called  the  Discriminant  of  the  quadratic. 


THE  THEORY  OF  QUADRATIC  EQUATIONS      139 

Example   1.     Determine  the  character  of  the  roots  of 

2a2-5a-18=0. 
Solution  :  1.    62  -  4  ac  =  ( -  5)2-4(2) (-  18)  =25  +  144  =  169  =  132. 
2.   By  Rule  1,  the  roots  are  real,  rational,  and  unequal. 

Example   2.     Determine  the  character  of  the  roots  of 

3^  +  2a;  +  l  =  0. 
Solution  :  1.    b*  -  4  ac  =  4  -  4  .  3  •  1  =  4  -  12  =  -  8. 
2.    By  Rule  3,  the  roots  are  imaginary. 

Example  3.     Determine  the  character  of  the  roots  of 

4a2-12a;  +  9  =  0. 
Solution  :  1.    62  -  4  ac  =  144  -  4  .  4  .  9  =  144  -  144  =  0. 
2.   By  Rule  2,  the  roots  are  real  and  equal. 

Note.  This  type  is  most  easily  understood  if  the  quadratic  is  solved  hy 
factoring.  This  example  becomes  (2x  —  3) (2*  —  3)=0.  The  roots  are  then. 
§  and  §.    It  is  customary  to  say  that  the  roots  are  equal. 

EXERCISE  61 
Determine  by  inspection  the  character  of  the  roots  of : 

1.  Gar*  +  7a;-5  =  0. 

2.  4z2-20x  +  25  =  0. 

3.  3z2-8z  +  5  =  0. 

4.  x2-9x-\-15=0. 

5.  5i*  +  7r  +  3  =  0. 

6.  9s2-l  =  12s. 


7. 

5  m  —  2  =  4  m2. 

8. 

4  y2 — y  =  6- 

9. 

5*2  +  7  =  8*. 

10. 

20^-41^  +  20  = 

=  0. 

11. 

7a?  +  3x  =  0. 

12. 

16  m2- 9  =  0. 

XIII.     EXPONENTS 

114.  In  the  preceding  chapters,  only  positive  integers  have 
been  used  as  exponents.  The  fundamental  definition  when  m 
is  a  positive  integer,  is : 

am  =  a  •  a  •  a  •••  a     (m  factors).  (§  3) 

115.  There  are  five  fundamental  laws  of  exponents.  When 
m  and  n  are  positive  integers : 

I    Multiplication  Law.        Just  as    a5  x  a7  =  aP, 

so  am  X  an  =  am+n. 
Proof  :   1.    am  =  a  •  a  •  a  •••  a     (tn  factors).  (§  114) 

2.  an  —  a  •  a-  a  •••  a     (n  factors).  (§114) 

3.  .•.  am  •  an  =  {a  •  a  •  a  •••  a  («•  factors)}  •  {« •  a .  a  •••  a  («  factors)} 

=  a  •  a  •  a  •••  a  {(m  +  n)  factors}. 

4.  .•.  an  •  aM  -  a"l+n.  (§  H4) 

II.   Division  Law.     Just  as   a9  -^  a*  =  a5, 

so  am  -=-  an  =  am-n.     (m  greater  than  n.) 

Proof:  1.    qm  =  ^'  ^'  ^  '  4  "'  &  '  a  '  a  '"  a  (^  factors) ,  (§114) 

an  fa  •  ^•ft-jk  •••  ft  {n  factors) 

2.  =  a  •  a  •••  a  {{m  —  n)  factors}  as  am~n.  (§  114) 

3.  .*.  am  ■«-  an  =  am_n. 


III.  Power  of  a  Power.     Just  as    (a5)3=  a15, 

so  (am)n  =  amn. 
Proof  :  1.    (am)n  -  am  ■  am  •  am  •••  am    (n  factors)  (§  114) 

2.  = :  am+m+m+  •••  +m    (n  terms).  (Law  Ij 

3.  .*.   {am)n  =  amn    {since  m .+  m  +  •••  +  m  (w  terms)  as  raw} 

IV.  Power  of  a  Product.      Just  as  (ab)5  =  a5b5, 

so  (ab)n  =  anbn. 
140 


EXPONENTS  141 

Proof  :  1.    («&)•  =  (ab)  -  (ab)  ■  (g&)  •••  (ab)     (n  factors)  (§  114) 

2.  =  {a  •  a  •  a  •••  a  (n  factors)}  •  {b  •  b  •  6  •••  b  (w  factors)}. 

(§  H4) 

3.  .\  («6)n  =  an  •  bn. 

V.   Power  of  a  Quotient.        Just  as     -  ]  =  -, 

-tJK-fr 

'"-'•(!)■-{*)(!)©••(;)  <■*■*->     """ 

„  _  q  •  g  •  q  •••  g  (n  factors) 

6  •  6  •  6  •••  b  (n  factors) 

3.  ,(f)"  =  ^.  (tiu) 

Involution  is  the  name  given  to  the  process  of  finding  a  power 
of  a  number.     (Compare  with  §  3.) 

EXERCISE  62 
Find  the  results  of  the  indicated  operations  in  the  following 
examples,  using  the  five  laws  above;   the  literal   exponents 
denote  positive  integers. 


1. 

x™-x. 

13. 

x15  -j-  a^3. 

25. 

O6)4- 

2. 

m12  •  m11. 

14. 

xl2  +  x». 

26. 

G/5)7- 

3. 

y5  -  yn. 

15. 

ifn  -f-  yn. 

27. 

(m4)8. 

4. 

m2a  '  ma. 

16. 

m',c  -+-  me. 

28. 

(-  a5**8)5. 

5. 

a3n  •  a2n. 

17. 

a4n  -+-  an. 

29. 

(j/Vw)4. 

6. 

br+1  •  b2. 

18. 

br+4  +  b2. 

30. 

(msnyy. 

7. 

cn~4  •  c5. 

19. 

cn+5  _j_  J 

31. 

«r. 

8. 

d2r+1  •  dr. 

20. 

rf2r+3_j..^ 

32. 

(6-)2. 

9. 

zr+1  •  zr~l. 

21. 

ar^H-ar* 

33. 

(_Cndm)*. 

10. 

tn-2  .  tn+S^ 

22. 

*»+•-*.«•-*. 

34. 

(a?y)m. 

11. 

wm+n  •  wm~n. 

23. 

ww+n  _j_  ?y»»-n# 

35. 

(rV)'. 

12. 

gn-r+1  .  grt 

24. 

gln-r+l  +  gr. 

36. 

(xmyny. 

142  ALGEBRA 


» ©'•     "■  (-?)'■     "  (5)" 


39. 


©'•      -  (9*-      » (£)' 


116.  Only  cube  and  square  roots  have  been  considered  in 
the  preceding  chapters.  More  general  roots  occur  in  mathe- 
matics. 

117.  Just  as  -y/x  indicates  the  cube  root  of  x  (§  3),  so  Vx 
indicates  the  nth  root  of  x. 

n  is  called  the  Index  of  the  root. 

The  nth  root  of  x  is  the  number  whose  nth  power  equals  x; 

thatis>  (Vxy=x. 

Thus,  y/x&  =  xs,  since  (a8)4  =  xn. 
y/%™  =  a;4,  since  (a4)5  =  x20. 
V-  x^y21  =—x2ys,  since  (—  se2y3)7  =  -  xuy21* 
The  number  under  the  radical  sign  is  called  the  Radicand. 

Rule. — To  find  the  nth  root  of  a  perfect  nth  power,  divide  the 
exponent  of  each  factor  of  the  radicand  by  n. 

Every  number  has  n  nth  roots.  Unless  something  is  said  to 
the  contrary,  the  principal  root  is  denoted  by  the  symbol  -y/~. 
If  n  is  even,  this  root  is  the  positive  root ;  if  n  is  odd  and  the 
radicand  is  negative,  this  root  is  negative. 

Evolution  is  the  name  given  to  the  process  of  finding  the 
root  of  a  number.     (Compare  with  §  3.) 

Historical  Note.  A  symbol  for  extracting  a  root  did  not  appear 
until  the  fifteenth  century.  In  Italian  mathematics,  the  first  letter  of  the 
word  Radix,  meaning  the  root,  was  used  to  indicate  the  square  root :  thus, 
R.  Presently  there  were  used  R.2a,  R.3a,  etc.  to  indicate  the  square, 
cube,  and  other  roots.  Chuquet,  a  French  mathematician  of  about  1500, 
used  R2,  R3,  etc# 


EXPONENTS  143 

In  Germany,  a  point  was  placed  before  a  number  to  indicate  that  its 
square  root  was  to  be  taken.  Two  points  were  used  to  indicate  the  fourth 
root,  and  three  the  third  root.  Keise,  1492-1559,  replaced  the  point  by 
the  symbol,  ^/.  to  indicate  the  square  root,  and  Rudolph,  1515,  used  the 
symbol,  y/^/,  for  the  fourth  root.  Stevin,  1548-1620,  used  the  better 
symbols:  V®»  V(D>  etc-  Girard,  1590-1632,  used:  fy  fy  etc.  Des- 
cartes used  the  vinculum  to  indicate  what  numbers  were  affected  by  the 
root. 

EXERCISE  63 


11.  ^64  a6b«.  21.  VoF. 

12.  </62o  a*b\  22.  J/aF. 

13.  -\/-27mV.  23.  -tybV. 

14.  -ty- 32  m*nw.  24.  i/-x"y™ 

15.  J/STf?.  25.  -Vtf^JF*. 

16.  -y/-b21c7du.  26.  y^V*. 


4jm4  _       V" 


:)ei 

bermine : 

1. 

Vs. 

2. 

V-it. 

3. 

V-ML 

4. 

</Sla\ 

6. 

^243  b5. 

6. 

■</mMn\ 

7. 

8. 

J-m* 

9 

V8 

17-  V*-  27-  Vs? 

18-  \-^r-  28-  V^ 


19     V-^3T-         29'    V^T 


10.    \j --.  20.    \r^rs-  30.    \  — 


118.  Fractions,  zero,  and  negative  numbers  are  used  as  ex- 
ponents. Up  to  this  point  the  symbols  a-3  and  a$  do  not  have 
any  meaning,  for  the  base  a  cannot  be  used  as  a  factor  minus 
three  times  or  two  thirds  times.     (See  §  114.) 


144  ALGEBRA 

119.   Meaning  of  a  Fractional  Exponent.     If  a1  is  to  obey  the 

multiplication  law  (§  115),  then  a?  •  a*  •  a?  =  a%  =  a2. 

.-.  (as)3=a2,  or  a's^v^ 

This  fact  suggests  the  definition  :  in  a  fractional  exponent, 
the  denominator  denotes  the  principal  root  (§  117)  of  the  power  of 
the  base  indicated  by  the  numerator.     In  symbols, 

m 


Thus:  x*  =  \/xs;    (-27)3=  V-27=-3. 

EXERCISE  64 

Express  with  radical  signs  and  find  the  values  of: 

1.  4*.  4.    32*.  7.    (-125)1  10.    (a6)*. 

2.  27*.  5.   81*.  8.    256*.  11.    (j/12)*.      - 

3.  (-8)*.  6.    64*.  9.    (-1000)*.  12.    (z10)*. 

13.    (-64a>V)*  14.    (32a5620)i  15.   (81  a;8?/4) *. 


Express  with  radical  signs: 

16.  2*  18.    5i  20.    4a;l        22.    2  abK      24.    mM. 

17.  A  19.    (4<c)*.      21.    3i/i        23.    (2abf.   25.    8  aM. 

Express  with  fractional  exponents : 

26.  Va\       28.    -V2a.      30.    ^S?.       32.    2^n~2.      34.    3yv^?. 

27.  \/?.       29.    2\/o!       31.    vW.      33.    4\ty«.      35.    a2  ^ 

120.   Meaning  of  a  Zero  Exponent.     If  a0  is  to  obey  the  multi- 
plication law  (§  115),  then  am  •  a0  =  am+0  =  aw. 
.•.  a°  =  am-r-am=  1. 


EXPONENTS  145 

This  partially  suggests  the  definition  :  the  zero  power  of  any 
number,  except  zero,  is  J. 

Thus:  50  =  1;   & .as  1  j    (-65)0  =  1. 

121.  Meaning  of  a  Negative  Exponent.     If  a~m  is  to  obey  the 
multiplication  law  (§  115),  then  a~m  •  am  =  a-m+m  =  a0  =  1. 

This  suggests  the  definition :  arm  =  — 

am 

Thus: 

ar*=:i.  «fi  =  ±=J—   f-lV4-      1      =-i-  =  ^-  =  -2 

EXERCISE  65 
Express  with  positive  exponents  and  find  the  values  of: 

1.  3~2.  5.    3"1  •  2"4.        9.    64  •  4"3.        13.    64~*. 

2.  2"3.  6.    5°.4"2.        10.    16~*.  14.    (-125)-*- 

3.  3"3.  7.   9-6"2.         11.    (-27)"*.      15.    (-32)"* 

4.  7°.  8.    100- 5"2.     12.   81"i 
Write  with  positive  exponents  : 

16.  a?b-5.  18.    2a~3.  20.   3  a~2b\ 

17.  (2  a)"3.  19.    (3a)-26V  21.    2-2m5n~4. 

22.   4a-66"3.  23.    (3  a)3  •  (3  6)"2. 

122.  Negative  Exponents  in  Fractions. 

1  t    2 

Example  1.     ?^"  =  ffl!.  =  <  .^^ 

z~*  1         af     1       a?1 


This  example  makes  it  clear  that  a  factor  may  be  trans- 
ferred from  one  term  of  a  fraction  to  the  other  provided  the 
sign  of  its  exponent  be  changed. 


146 


ALGEBRA 

Example  2. 

5  ^y~H~z 
w-H 

_5#2w?2 
ty4z3  ' 

Example  3. 

3a2--3a26c-1d"3. 

cds 


EXERCISE  66 
Write  with  positive  exponents : 


„,3  *     O  „,-5  *         7  ,,-*  ". 


2/3  2y~&  "p 


tfy 


2    ^  £  6    3  a"3&2  8.  5a*b  - 

'  6"3#  *   5- (2 1/) -3'        '   2c-2d4'  6A 

Write  without  any  denominator : 


2/2 


&* 


15. 


,-6 


**  2  <r2d* 

a~*  ..  „    4  a^mt 


raw  14.   _ i6. 


10'    c4'        .     12'      eZ3    '  '    b~*c*  '   2b57i-$ 

Historical  Note.  In  the  note  following  §  14,  credit  is  given  to 
Herigone  for  having  grasped  the  idea  of  an  exponent,  and  for  introducing 
a  rather  good  notation.  As  early  as  1484,  another  French  mathematician, 
Chuquet,  had  had  some  idea  of  an  exponent  and  had  written  expressions 
involving  a  form  of  negative  exponent  and  also  the  zero  exponent.  His 
ideas,  however,  did  not  spread  far.  Other  attempts  to  introduce  general 
exponents  were  made  between  that  time  and  the  time  of  Newton.  To 
Newton  must  be  given  the  credit  for  having  finally  fixed  the  present  form 
of  writing  the  various  kinds  of  exponents. 

123.     The  Fundamental  Laws  for  Any  Rational  Exponent. 

The  symbol  xn  has  been  defined  now  (§§  114,  119,  120,  121) 
for  all  rational  (§  112)  values  of  n.  The  five  fundamental  laws 
which  have  been  proved  for  positive  integral  exponents  (§  115) 
apply  also  for  other  rational  exponents.  This  fact  will  be 
assumed  without  proof  in  this  text. 


EXPONENTS  147 

EXERCISE  67 
Law  I 

Example,     a7 .  a~5  •  a° .  a?  =  a~5+0+h  =  aK 

1.  Express  Law  I  in  words. 

2.  Multiply  each  of  the  following  numbers : 

r3;       r7;       s~4;       rV;       r~V"6;       f*jP 
by:  (a)  r5;    (b)  r"6;    (c)  s3;    (d)  rV;    (e)  r-4s~a. 

3.  Multiply  each  of  the  following  numbers  : 

**j       a?*;       y*>       ajV>       *Yi 
by:  (a)  «*;    (6)  «£;    (c)  yt;    (d)  xty. 

4.  Multiply  each  of  the  following  numbers  : 

m ~»;       n  *;       m^"17;       irti«j       m~*w ~*. 
by:  (a)  m;     (6)  m%;     (c)  w~*;     (cQ  m"V. 

Multiply : 

5.  a*  +  ah*  +  6^  by  a*  —  $. 

6.  2a-1-7-3aby4a-1  +  5. 

7.  of* +  2af*  +  4aT*  +  8  by  af*-2. 

8.  a?*  -|-  a;^  4-  y*  by  a;^  —  B*y«  +  y^. 

Find : 

9.    (a~*  +  &*)(flf*-&*).  12.  (»*  -  6)  (»*  + 13). 

10.  (afi  - 1/"^)2.  13.  (r*  -  sty. 

11.  (**-*•)»  14.  (a*  +  7  6"1)  (a*  -  8  &-1), 


148  ALGEBRA 

Law  II 

Example,     m2*  -s-  m~*  =  m2*  ~ ( ~    =  m  . 

15.  Express  Law  II  in  words. 

16.  Divide  each  of  the  following  numbers : 

by:  (a)  *3;    (6)  r4;    (c)  *j    (d)  rl 

17.  Divide  each  of  the  following  numbers : 

c-5d4;   <fd«;   c*cT*;   cV*; 
by:  (a)  cd;  (b)  c~2d-\ 

18.  Divide  a-3  4-  a"2  +  a-1  by  or4. 

19.  Divide  4  or6 +  6  or4 +  12  or2  by  2  or2. 

20.  Divide  a4  +  a3  +  a2  4-  a  by  a*. 

21.  Divide  a2  +  62  by  a*  +  b$. 

22.  Divide  a  —  1  by  a?  4  1. 

23.  Divide  a  —  4  a*  4  6  a^  —  4  a^  -|- 1  by  a^  —  2  a*  4- 1. 

Law  III 
Example,     (of  *)*  =  of  2 '  $  =  af». 

24.  Indicate  and  find  the  values  of  the  following  numbers : 

«p;  Or12)";  (*8)-;  C^)n;  (*2-4)w; 

when  71  is:  (a)  2;   (6)  -3;    (c)  i;    (d)   -J;   (e)   -  f . 
Laws  IV  and  V 
Example  1.     (x~5y*)~*  =  a;"5 ' "  V '  ~*  =  #*/"*• 


/r-i\-2      r-3--2       *3 
Example  2.     (M    ='__  =  ^  =  ^. 


EXPONENTS  149 

25.  Express  Law  IV  in  words. 

26.  Express  Law  V  in  words, 

27.  Indicate  and  find  the  values  of : 

(a2&"3)n;    (m-3^)";    (af^"*)";    (r~as-b)n$ 
when  n  is:  (a)  2;  (6)   -4;  (c)  i;  (d)   -> 

Eind  the  values  of: 

28.  (-8)1 

Solution  :  (-  8)*  =  [(-  8)^]2  =  [\/^8]2  =  (-  2)«  =  4. 

29.  25i  32.    81*.  35.    (4aJ2)i  38.    (—32)*. 

30.  9*.  33.    49*.  36.    (243  ar5)!      39.    (64a?V2)i 

31.  8*.  34.    (-27)*.       37.    (16  m4)  I      40.    (-125)1 

41.  (_64a366)l  43.    (256  afy8)*. 

42.  (-128  m7)*.  44.    16L25. 

„K     q-      !•*     10L5xl02 

45.  Simplify  - • 

r      *  10125 

Solution  :  101-5  x  102  _  10W+2_i.2s  _.  10&25# 

1Q1.25 

46.  Multiply  each  of  the  following  numbers : 

10175.     -,02.23.      103.47.      }(£».      1()9.86 . 

by  (a)  10;  (6)  100;  (c)  101-25. 

47.  Examine  the  results  of  46  (a)  and  (6).  What  is  the 
effect  upon  the  exponent  of  a  power  of  10  when  the  power  is 
multiplied  by  10?  by  100? 

48.  Replace  the  word  "multiply"  in  Example  46  by 
"divide"  and  solve  the  resulting  exercises. 

Simplify : 

r,2m— 3n       rl—5m—n  nn  .   ( nn—\\n 

49.  - 50.  ^      >  • 

^TO-^  /7M+I  .  /?*»-! 


XIV.     RADICALS 

124.  A  Radical  is  a  root  of  a  number  indicated  by  a  radical 
sign ;  as,  V5,  Va,  V#  +  1. 

If  the  indicated  root  can  be  obtained,  the  radical  is  a  rational 
number;  if  it  cannot  be  obtained,  it  is  an  irrational  number 
(cf.  §  112).   ' 

125.  The  index  (§  117)  determines  the  Order  of  the  radical. 
Thus,  Vx  +  1  is  a  radical  of  the  third  order. 

126.  An  introduction  to  radicals  of  the  second  order  (square 
roots)  has  been  given  in  Chapter  VII.  In  §  69,  a  means  of 
simplifying  radical  expressions  in  order  to  find  their  approxi- 
mate values  is  illustrated.  Some  methods  of  simplifying  more 
complicated  radical  expressions  will  be  given  in  this  chapter. 
These  methods,  like  the  one  in  §  69,  lead  to  more  economical 
and  often  to  more  accurate  methods  of  finding  the  approximate 
arithmetical  values  of  the  expressions. 

It  will  be  of  interest  also  to  find  that  radicals,  like  integers 
and  fractions,  can  be  added,  subtracted,  divided,  etc. 

127.  Radicals  of  the  second  order  will  be  emphasized. 
Where  the  final  expression  involves  only  square  roots  of  arith- 
metical numbers,  the  approximate  arithmetical  value  should  be 
found  as  in  the  examples  solved  in  the  text. 

128.  Two  principles  are  used  frequently  in  this   chapter: 

(A)  (^/x)n  =  x  (§  117).  Thus,  (V2)2  =  2.  From  this  it 
follows  that  V22  =  2.     Similarly  -\/3~5  =  3. 

(B)  Vab  =  -Va  •  -Vb.     Thus,  -^7^9  =  jf?  -  ^9. 

160 


RADICALS  151 

This  principle  may  be  expressed :  the  nth  root  of  the  product 
of  two  numbers  is  equal  to  the  product  of  the  nth  roots  of  the 
numbers. 

REDUCTION  OP  A  RADICAL  TO  ITS  SIMPLEST  FORM 

129.  Reducing  a  Radical  to  a  Radical  of  Lower  Order. 

Example  1.     ^125  =  -y/Wm  (5s)*  =  5'  =  5*  =  VS", 
.-.  -5/125  =  V5  =  2.23+. 

Example  2.       -^64  =  (26)*  =  2*  =  2*  =  -tf?  =  ■$/& 

.*.  the  ninth  root  of  64  may  be  found  by  obtaining  the  cube 
root  of  4.  In  the  chapter  on  logarithms,  a  method  for  deter- 
mining a  higher  root  of  any  number  will  be  given. 

EXERCISE  68 
Reduce  to  radicals  of  lower  order ;  see  §  127 : 

1.  -V25.  7.  -v/16.  13.  ^64.  19.  -^125^. 

2.  VWO.  8.  -fyEL  14.  K/l.  20.  ^32^?. 

3.  \/8.  9.  ^32.  15.   \/216.  21.  ^81^ 

4.  -^36.  10.   \/49.  16.  ^100.  22.  ^8^ 

5.  ^27.  11.  -y/25.  17.  ^243.  23.  -y/21  ctoP. 

6.  ^343.  12.  fy§.  18.  ^121  dlb\  24.  ^256  aV. 

130.  Removing  a  Factor  from  the  Radicand. 

Example  1.     V75  =  V25^3  =  V25  •  V3  =  5  .  V3. 

...  V75  =  5(1.732+)  =  8.66+.     (See  also  §  65  ) 

Example  2.     <^96  a5b12<?=  ^32^W  .  y/Wf 

=  2ab2c</3W. 


152  ALGEBRA 

Rule.  —  To  simplify  a  radical  by  removing  factors  from  the  radi 
cand: 

1.  Resolve  the  radicand  into  two  factors,  the  second  of  which 
contains  no  factor  which  is  a  perfect  power  of  degree  corresponding 
to  the  order  of  the  radical. 

2.  Find  the  required  root  of  the  first  factor ;  multiply  it  by  the 
indicated  root  of  the  second  factor. 


Simplify  by  removing  factors  from  the  radicand ;  see  §  127 : 
1.   V52.  5.  V98.  9.  V125.  13.   a/4003". 


2    V90.  6.   V96.  10.  V99a2.         14.  V54m. 


3.  V80.  7.  V112.  11.  V60ary.      15.  V375Z6. 

4.  V63.  8.  V108.  12.  V200mV.  16.  -fylWa*. 

17.  -\/l28a*/4.  19.  </162.  21.   J/MW. 

18.  ^1125 m¥.  20.  </64a^.  22.  <7243 ny. 

23.  -\/12Sx&if.  26.  V27a36-36a262+12a63. 

24.  a/128 xy.  27.  V5  or*  +  30  a2  +  45  x. 


25.  V(a2-4  62)(a-2  6).  28.  V(^-»-6)(a?24-2a;-15). 

29.  M^M^W^M. 

^8       ^2*     -^23       2 

30.  <£2Z.      32.   x4/1^-       34'    4^  '        36.    jdZ. 
^27m6       '        \16a4  \32c5  \64z6 


31.    M£.  33.    Affi.        35.    V&         37.  V  — I 

\125  \81«*  ^ofy10  M28m7wM 

131.  Changing  a  Fractional  to  an  Integral  Radicand. 
Review  §  66  and  Exercise  27.  The  method  of  §  66  applies 
to  radicals  of  higher  order. 

s[27        sf    33-2a2'        3    3/__ 
Example.     ^_  =  ^  (__  =  —  V2  <A 


RADICALS  153 

Rule.  —  To  change  a  fractional  to  an  integral  radicand : 

1.  Multiply  both  numerator  and  denominator  of  the  fraction 
by  such  a  number  as  will  make  the  denominator  a  perfect  power 
of  degree  corresponding  to  the  order  of  the  radical. 

2.  Simplify  the  resulting  radical  as  in  §  130. 

EXERCISE  70 
Express  with  integral  radicand  s;  see  §  127: 

■■>!   >-m  "M-  -& 

-4    '-4    »4     »# 

-4    »■#_    *C    -l? 
*V£-  »*?  »#  -xSI-  . 

,vn-  »^-  -^  «y£® 

132.    To  Introduce  the  Coefficient  of  a  Radical  under  the  Radical 
Sign. 
Example.   2aty^=-V(2a)'3  -  •v/3^='v/8a3 .3«2=-\/24aV 

Rule.  —  To  introduce  a  factor  under  the  radical  sign  : 

1.  Raise  the  factor  to  the  power  denoted  by  the  index. 

2.  Multiply  the  radicand  by  the  result  of  step  1. 


154  ALGEBRA 

EXERCISE  71 
Introduce  under  the  radical  sign  the  coefficients  of : 

1.  5V2.  4.  5\/4.  7.  4aV8a.  10.  a?y2J/tfy\ 

2.  8V3.  5.  2^5.  8.  7ajV6a?.       11.  3m*</2m. 

3.  4^/5.  6.  3^2.  9.  3od\/5c?.      12.  2a-\/Ttf. 

13.   (l+a)^=i.  15.   «I^J«±L 


14.  (^-1)^-^+1.  i6.  Stpis/1 


2x 


(x  +  iy 


133.  Similar  radicals  are  radicals  which,  in  their  simplest 
form,  do  not  differ  at  all  or  differ  only  in  their  coefficients ; 
thus,  2v  asc2  and  3^/ax?  are  similar  radicals. 

134.  Addition  and  Subtraction  of  Radicals.  Review  §  69  and 
Exercise  28.  The  methods  of  §  69  apply  to  radicals  of  a 
higher  order. 

Example.    -s/\  -  V%i  +  ^54  =  ■#}  -  ^873  +  -^27^2 
=  \3/i  -  2^3  +  3^2  m  31^2  -  2.^3. 

Rule.  —  To  add  or  subtract  radicals : 

1.  Reduce  them  to  their  simplest  form. 

2.  Combine  similar  radicals  (see  §  69)  and  indicate  the  addi- 
tion or  subtraction  of  those  which  are  dissimilar. 


EXERCISE  72 
Simplify  the  following  expressions  ;  see  §  127 : 

1.  V98-V32.  4.  i/M+</T6.  7.  ^32--J/l62. 

2.  2 \/80  +  V180.      5.  ^192m -  y/Sm.     8.  4*64 -  y/2. 

3.  3V24-V15U.      6.   y/2Xl&  +  ^24ril      9.   ^/3  +  ^/l92. 


RADICALS  155 

10.  m*V&m*  +  m-5^108  m5  -  -^500  m*. 

11.  arVl50 x  +  V96l? -  VSSa?  -  a;V24a?. 


12.  Vf+VV-  ,       5c 

13.  V^  +  Vf. 


16 


14.  V| 


+V*  18.  ^?+</A. 

is.  Vf  +  VA-Vf. 

135.   Reduction  of  Radicals  of  Different  Orders  to  Equivalent 
Radicals  of  the  Same  Order. 

Example.     Reduce  V2,  V2,  and  V5  to  equivalent  radicals 
of  the  same  order.     Determine  which  is  the  greatest  number. 

Solution  :    1.    By  §  119 ,  V2  =  (2)  *  =  2^  =  y/&  _  '^/gj. 

2.  ^3  =(3)*  =  *&  =  ^  =^8l., 

3.  #>  =(5)*  =  ffA  -^  =1-^l2S. 

4.  .*.  y/Z  is  the  greatest  number. 

Rule. — To  reduce  radicals  to  equivalent  radicals  of  the  same 
order : 

1.  Express  the  radicals  with  fractional  exponents. 

2.  Reduce  the  exponents  to  a  common  denominator. 

3.  Rewrite  the  resulting  expressions  with  radical  signs. 


156  ALGEBRA 

EXERCISE  73 
Reduce  to  equivalent  radicals  of  the  same  order: 

1.  V3  and  V5.  6.    y/xy,  -i/yz,  and  -fyxz. 

2.  V2  and  y/S.  7.    y/Za,  ^26,  and  Vo^". 

3.  -tyM  and  v^V  8.    %%  V8,  and  Vl3. 

4.  V2  and  Vl2.  9.    Vl^a  and  y/T+x. 

5.  V4 and  V&  10.    VcT+6  and  V^^. 
Arrange  in  order  of  magnitude : 

11.  -y/2  and  -y/3.  14.    V3  and  Vl5. 

12.  y/H  and  VS.  15.    V3,  ■&$,  and  </?. 

13.  VlO  and  y/l.  16.    Vli,  V6,  and  y/TtB. 

MULTIPLICATION  OF  RADICALS 
136.   Multiplication  of  Radicals  of  the  Second  Order. 

Example.     2V3-  V6=2  V^=2V^2=2 -3  V2=6  V2 
.-.  2  V3  •  V6  =  6  V2  =  6  (1.414+)  =  8.484+. 

EXERCISE  74 
Find  the  products ;  see  §  127  : 

1.  V2-V10.  4.    V5-VT5.  7,   2y/5-3y/B. 

2.  V3-VI2.  5.   2V3V21.         8.    (3V3)2. 

3.  VT-Vli.  6.   3V20-V10.      9.    (5V2)2. 
10.    (2V7)3.  13.    Vse  +  1-Vs^L 
11     5y/6x.2y/3x.  14.    (Va^5)2. 


12.   3V3m2.2Vl5m.  15.   (3Va  +  2)2. 


RADICALS  157 

16.  Multiply  2  V3  +  3  V2  by  3  V3  -  V2. 
Solution  :  2  VS  +  3  V2 

3V3-V2 
18  +  9V6 

-2V6-6 
18  +  7\/6-6=  12  +  7V6. 
A  (2V3  +  3V2)(3V3  -  V2)=  12  +  7(2.44)=  12  +  17.08  =  29.08. 

Find  the  following  products  : 

17.  (5-V3)(5+V3).  24.  (V2  -  7)(V2  +  7). 

18.  (2a-V&)(2a+V&).  25.  (2  V3 -f  5)(2V3 -5). 

19.  (V3  +  7)(V3-8).  26.  (Va  -  V&)(Va  +  V5). 

20.  (2  +  3V3)(6-V3).  27.  (V^+T+l)2. 

21.  (V2-4)(3V2-5).  28.  (Va^3-4)2. 


22.  (4-f  V£)2.  29.    (Va-Va;  +  5)2. 

23.  (2-3V7)2.  30.    (V^+l-V^l)2. 

137.   Multiplication  of  Radicals  of  Any  Order. 
Example  1.    -ty±&  .  VWrf  =  tyWs?  =  2 xVx. 
Example  2.    V2^ .  ^TciF  =  Vj2df  •  ^4^*7 


=  V23-a3-42.a4=  V23-24.a7 

=  2a^/2^".  t 

Rule.  —  To  multiply  monomial  radicals : 

1.  Reduce  the  radicals,  if  necessary,  to  equivalent  radicals  of  the 
same  order.    (§135.) 

2.  Multiply  together  the  radicands  obtained  in  step  1  for  the  radi 
cand  of  the  product ;  place  it  under  the  common  root.    (§  128,  B) 

3.  Simplify  the  result  of  step  2  as  in  §§  130  and  131. 


158  ALGEBRA 

EXERCISE  75 

Find  the  products : 

1.  </!•  -y/2.  8.    Va-Va. 

2.  2^3.^18.  9.    -y/V'tyb. 

3.  5-^9^  •  -y/Ss?.  10.    Vm.\/»?. 

4.  </9.-^27.  11.    ^2.^8: 

5.  "V^^V^.  12.     VlO-A/T 

6.  \/16a?  i  -#12a?.  13.    ^9a*  ■  Vl5a". 

7.  ^{fS^f'-VmjF.  14.   -v^?.-^ft 

15.   5  yjm%n  •  V6  ra3?i6. 

DIVISION  OF  RADICALS 
138.   Division  of  Monomial  Radicals  of  the  Same  Order. 

Example  1.      V6  --  V2  =  Jf  =  V3.     .\  V6-J-  V2 =1.732+. 

.-.  V12+- V 15  -  2(2'236+)  =  igg  =  .894+ 
5  5 

Examples.   ^-,^18  =  ^=^=^  =|^/12. 

Rule.  —  To  divide  monomial  radicals  of  the  same  order. 

1 .  Divide  the  radicand  of  the  dividend  by  the  radicand  of  the 
divisor,  and  write  the  result  under  the  common  radical  sign. 

2.  Simplify  the  result  as  in  §§  130  and  131. 


RADICALS  159 

EXERCISE  76 

Perform  the  indicated  divisions;  see  §  127: 

1.  V8-5-V2.  9.   4cVl2~c^-j-V2^. 

2.  VU-I-V7.  10.  yi5-j-V£. 


3.  Vl2 m  -v-  v4m.  11.  11  Vxy -*-  V^ y. 

4.  6V15-J-2V5.  12.  V?f-*-V2 

5.  2a\/72-*-aVl8.  13.  Vl0 -*- V6. 

6.  V2-VJ.  14.  V12-J-VT. 

7.  15V^"3^-5V^&.  15.  V33--VI5. 

8.  6 V187^-j- 2V6^.  16.  (8VI2-6V3)^-2V3. 

17.  (15V2^+25V6^)-*-5V2^. 

18.  (V8  +  2VlO)-5-V3. 

19.  (3VI5-4VT8)W6. 

20.  -^135-^-^5.  25.   2a-y/U^-i-a^/6xf. 

21.  v^-h-v^.  26.   3aby/^-*-aVm/& 

22.  \/26l?--^39al  27.   6m2^<^-2mri\/aF. 

23.  •v/i92^-j-</3^.  28.    V5W-hV?F. 

24.  \/l2oa262^-v/256^.  29.    V^-5- V^". 

139.   Division  of  Monomial  Radicals  of  Any  Order. 

Example  1.     A  =  ^  =  ^7|  =  ^. 
-^4      -^i      v4 

Example  2.     _  =  _  =  ^_  =  ^  =  x/-  =  -V243. 


160  ALGEBRA 

Rule.  —  To  divide  one  radical  by  another : 

1.  Reduce  the  radicals,  if  necessary,  to  equivalent  radicals  of 
the  same  order. 

2.  Divide  the  radicand  of  the  dividend  by  the  radicand  of  the 
divisor  for  the  radicand  of  the  quotient  and  write  the  result  under 
the  common  radical  sign.     Simplify  the  result. 

EXERCISE  77 
Find  the  quotients : 

1.  2-*-v/2.  4.    5-^25.  7.    VR-f-'^L 

2.  3---\/3.  5.  d+</a.  8.   VS+V9. 

3.  3  +  -WT.  6.   V2-*-^2.  9.    y/Wi-*-yWz. 

10.  \/t2€?^^/2E  13.    ^T+.A/f. 

11.  \/J-*V$  14.    -^20^-^125. 

12.  a/SI^^-v7^/.  15.    \^+^a^. 

140.    Division  by  a  Binomial  Quadratic  Surd. 
Example  1. 

1  (2-V3)  ^2-V3  =  2-V3| 

2+V3      (2  +  V3)(2-V3)        4-3    "         1 
...  l--(2+V3)  =  2-1.732+=.267+. 

Note.  2  +  V3  is  multiplied  by  2  —  V.3,  thus  giving  the  product  of  the  sum 
and  the  difference  of  two  numbers.  The  product  is  the  difference  of  their 
squares.    2+V3  and  2—  V3  are  called  Conjugate  Surds. 

In  general,  the  product  of  two  conjugate  surd  expressions  is 
a  rational  number,  for  (a  +  Vb)(a  —  V&)  equals  a2  —  (Vbf 
=  a2-b. 

Rule.  —  To  divide  a  number  by  a  binomial  quadratic  surd: 
1.    Multiply  both  dividend  and  divisor  by  the  conjugate  surd  of 
the  divisor,  and  simplify  the  result. 


RADICALS  161 

Example  2. 

3V2  +  1  =  (3  V2+  1)(2  V2  +  1)  =  12  +  5V2+1. 
2V2-1      (2V2-1)(2V2  +  1)  8-1 

3  V2  +  1  =  13  +  5(1.414+)  =  20.07+  =  9  g6+ 
'*'  2V2-1  7  7 

Note.  Since  in  this  method  of  division  the  original  fraction  is  changed 
into  an  equivalent  fraction  with  a  rational  (§  242)  denominator,  the  process  is 
referred  to  as  "  Rationalizing  the  Denominator." 

EXERCISE  78 

Perform  the  indicated  divisions;  see  §  127: 

1.   _?_.  4.   -J 7    3-V2, 

3-f-Vo  V3-4  '  4  +  V2 

2.  -1— .       5.  — *_.       8.  vg+». 

V6-2  3  +  2V5  Va-6 

3    _i_.  6.    2+V^.  9.    Vg-^. 

'  3-V5  '  1+V3  '   Vs  +  Vy 


io.  vs-vs.  is.  aaEIfi; 

V5  +  V2  Va-2+2 

2V2+3_  14    Va-6+Va 

3V2  +  2  Va  —  6  —  Va 

12  W2-fg  15  yr+^_yr=^t 

3V2-6  Vl+a  +  Vl-a 

141.   Involution   and  Evolution  of  Radicals   is   accomplished 
in  the  case  of  monomials  by  the  use  of  exponents. 

Example  1.     (-\/l2)3  =  (12*)3  =  12*  =  12*  =  Vl2  =  2  V3. 
.-.  (a/12)3  =  2  (1.732+)  =  3.464+. 

Example  2.     "v/( a/27^3)  - { (27  a8)*1*  =  5  (3  XY 1  *  =  (3  »$ 


162  ALGEBRA 

EXERCISE  79 
Simplify  the  following  expressions;  see  §  127: 

1.  (</5)2.  6.  (2aVb)\  11.  (V3a-2)\ 

2.  (-^8)2.  7.  {Vl~^f.  12.  (Vi8^y)3. 

3.  (-^128)8.  8.  (VW^)4.  13.  -^(VM). 

4.  (</6)4.  9.  (-^3)7.  14.  ^(V27). 

5.  (a/16)2.  10.  (5m^96<)2.  15.  V(-\/25). 

16.  -\/(^32^).  19.    ->/(</2^. 

17.  V(^49).  20.    V(</9~tf). 

18.  -v^(VlO).  21.    V(-v^»~6aj  +  9). 

142.  Square  Roots  of  a  Binomial  Quadratic  Surd.  It  is  possible 
to  find  the  square  roots  of  some  binomial  surds  by  inspection. 

(V2-V3)2  =  2-2V6  +  3  =  5-2V6. 

Notice  that  the  square  of  the  binomial  surd  is  a  binomial ;  that  5  is 
the  sum  of  the  two  radicands  2  and  3  and  that  the  radicand  6  is  the 
product  of  the  radicands  of  the  given  binomial.  This  example  suggests 
the 

Rule.  —  To  find  the  square  root  of  a  binomial  surd  (§  67)  : 

1.  Reduce  the  surd  term  so  that  its  coefficient  is  2. 

2.  Separate  the  rational  term  into  two  numbers  whose  product 
shall  be  the  radicand  obtained  in  step  1. 

3.  Extract  the  square  roots  of  the  two  numbers  of  step  2  and 
connect  them  by  the  sign  of  the  surd  term  (§  15,  c). 

Example.     Find  the  square  roots  of  22  —  3V32. 


Solution:  1.    V22  -  3V32  =  V22  -  V9  •  8  .  4  =  V22  -  2  V72. 
2.    22  =  18  +  4  and  18  x  4  =  72. 

8.  '.-.  V22-3V32  =  ± ( Vl8  -  VI)  =  ±(3V2  -  2).     (§  59). 
Check  :  (3V2  -  2)2  =  IS  -  12V2  +  4  =  22  -  3V16T2  =  22  -  3  V32. 


RADICALS  163 

EXERCISE  80 
Find  the  square  roots  of : 

1.  11  +  2V28.  4.   8-V60.  7.   9-3V8. 

2.  17-2V72.  5.    6+V32.  8.   8  +  4V3. 

3.  11-2V30.  6.   6-V20.  9.   20-6VH 

IMAGINARY  NUMBERS 

143.  An  introduction  to  imaginary  numbers  was  given  in 
Chapter  VIII.  Review,  if  necessary,  paragraphs  82  to  85 
inclusive. 

144.  Powers  of  the  Imaginary  Unit  i. 


By  §  82,  i  is  ^^V,  therefore  i2  =  -1.     (§  117.) 
i3  =  i2 .  t  =  (—  1)  i  —  —  i. 
£«=i8.t  =  (-*)*  =  -(«  =  -(-1)  =  1. 
i5  =  i4 .  t  =  1  •  i  =  i. 

Thus,  the  first  four  positive  integral  powers  of  i  are  i,  —  1, 
—  i,  and  1 ;  and  for  higher  powers,  these  numbers  recur  in  the 
same  order.     Find,  for  example,  i6,  i7,  and  i8. 

145.   Multiplication  of  Imaginary  Numbers. 
Example  1. 

■y/Z~2  .  V^3  =  iV2  .  iV3  =  iW6=  (-  1)  •  V6  =- VB. 
Note  that  each  number  is  expressed  in  terms  of  the  unit  i, 
and  that  the  fact  that  i2  —  —  1  is  used. 

Example  2.     Find  the  product  (2  -  V^)  (5  +  -V^-3). 
Solution  :  (2  -  V^3) (5  +  V^3)  =  (2  -  iVS) (5  +  i V3). 
2  -  iV3 
5  +  *V3 
10-5iV3 

+  2iV3-i2.3 


10-3iV3-(-  l)3  =  13-3z\/3 


164 


ALGEBRA 


EXERCISE  81 


Find  the  products : 

i.  V17!  •  V^9. 

3.    V^6  •  V^3. 


4. 


8.  a-V—b  •  cV—  6. 

9.  m  V— r  •  «V-  s. 

10.  V^^a-  -V7^. 

11.  (2+V^l)-(2-V^I). 

12.  (3  +  V:r5)(3-V^5). 

13.  (7  +  V^6)(7  +  2V^B). 

14.  (9-V^3)(ll+V^l3). 
16.    (4-V^5)2. 


5.  V^9a*  •  V-  16  a2. 

6.  2V:r3-3V^3. 

7.  5V^2-4V:^2. 
15.    (_l  +  V^3)(-l-V-3) 

i7.  («+yzy)(»Tvcry).  19.  si(_i-v"^3)i! 

18.    Si(-l+V^3)S2.  20.    Ji-(-l-V^3)i 

146.  Division  of  Imaginary  Numbers. 

Example  1.     V^  =  iVl?  =  V^L  V4  =  2. 
V-3       *V3       V3 

Example  2. 

10  10  10.?V2        10iV2 


V-2     tV2     i2-V2.V2         -2 


=  -5fV2. 


Example  3. 


2(1  -  ;V3) 


l  +  V-3     1  +  »V3      (l  +  *V3)(l-tV3) 
_  2(1  -  tV3)  _  2(1  -  tV3)  _  1  -  J  V3 


l-r-3 


1  +  3 


2 


Note.  As  in  division  of  real  radicals,  rationalize  the  divisor,  by  multiply- 
ing by  the  conjugate  imaginary.  Thus,  to  rationalize  3— V— 5,  multiply  it 
by  3+V^5;  the  product  will  be  3*  — (V^5)2,  0r  9  — (— 5),  which  is  14. 


RADICALS  165 

EXERCISE  82 


Find  the  quotients : 


1.  V-25-J-V-5. 

2.  vCT^+V—a 

3.  V^W^. 

4.  VSS-t-V^. 


5.    V—  ab  4-V— &c. 


6.  V—  a-^V—  a2. 

7.  2V^~T5W^3. 


11. 


12. 


13. 


v- 

-40  or* -5- 

2 

1- 

•  V^3 

14 

•  V^T 

1- 

V=I 

5  +  4V^6 

5- 

4V^6 

7- 

•  6V^3 

8.  12V-18-f-4V-2. 

14. 

9.  aV^I08-^-V-4.  3  +  2V-3 


147.  Application  of  Radicals.  In  Chapters  VIII  and  XI 
irrational  (§  124)  roots  were  found  for  quadratic  equations. 
Checking  by  substitution  in  such  cases  was  not  recommended 
at  that  time. 

Example.     Solve  the  equation  x2  +  x  —  1  =  0. 


Solution:   1.    By  the  formula  (§78),  x  = — 3  ^        +  4 

_  1  +  V5             -  1  -  V5 
2.  ...n  =  _±_;fl  =  _ 

Check:  Does        (~  1  ±  ^)2+  (~  1  ±  V5)  -  1  =  0? 

Does  1-2V5  +  5      -l-fV5_1  =  0? 

4  2 

Does  1-^5  +  5-2+^5-4  ^  0  ?    Y^ 


166  ALGEBRA 

EXERCISE  83 

1.  Check  the  second  root  r2  above  by  substitution. 
Solve  and  check  the  following  equations: 

2.  a?2- a;  — 1  =  0.  5.   x2  +  x  +  l  =  0. 

3.  a^-2a;-2  =  0.  6.  ar3  +  l  =  0.     (See  §  95.) 

4.  2/2-3#  +  l=0.  7.   a?-8  =  0. 

8.  In' a  higher  course  in  mathematics  (trigonometry)  certain 
six  numbers  occur,  five  of  them  bearing  the  following  indicated 
relations  to  the  sixth;  calling  the  numbers  s,  c,  t,  8,  C,  T: 

(a)  o  =  VI=?.         (c)   S— -L=.       («)  T=^^.. 

VI  — s2  8 

If  s  =  — =,  find  c,  £,  (Si  (7,  and  T7  in  simplest  radical  form. 

9.  If  s  =  ^— ,  find  c,  £,  S,  C,  and  T7  in  simplest  radical  form. 

Z 

10.  When  factoring  expressions  in  Chapters  II  and  IX, 
only  factors  involving  rational  numbers  were  permitted.  Fac- 
tor the  following  expressions,  using  irrational  or  imaginary 
numbers,  if  necessary : 

(a)  x*-2.  (d)  x>  +  2.  (g)  5x2-9. 

(b)  x*-5.  (e)   a^  +  4.  (h)  2x*-5. 

(c)  x>  +  9.  (/)   3^-4.  (f)   ax*-b. 

IRRATIONAL  EQUATIONS 

148.  An  Irrational  Equation  is  one  in  which  the  unknown 
number  appears  under  a  radical  sign  or  with  a  fractional 
exponent. 


RADICALS  167 

149  It  is  agreed  that  the  radical  sign  or  fractional  ex- 
ponent shall  denote  the  principal  root  (§  117)  ;  thus  the  square 
root  shall  always  denote  the  positive  root. 

150.  The  following  examples  illustrate  the  methods  of  solu- 
tion of  irrational  equations. 


Example  1.   Solve  the  equation  x  —  1  —  Vx2  —5  =  0. 


Solution:   1.    Transposing,      x  —  1  =  Vx2  —  5. 

2.  Squaring  both  members,  x2  —  2  x  +  1  =  x2  —  5. 

3.  .*.  -  2  x  =  —  6,  or  x  =  3. 

Check  :  Does  3  -  1  =  VP^B  ?    Does  2  =  VI  ?    Yes.     (See  §  149.) 

Note.  When  a  single  radical  occurs  in  an  equation,  transpose  the  terms 
until  the  radical  is  on  one  side  by  itself  and  the  remaining  terms  are  on  the 
other  side.  Then,  if  the  radical  is  a  square  root,  square  both  members  of  the 
equation;  if  it  is  a  cube  root,  cube  both  members. 


Example  2.   Solve  the  equation  x  —  1  -f  V#2  —  5  =  0. 


Solution  :  1.    Transposing,  y/x2  —  5  =  1  —  x. 

2.  Squaring  both  members,      x2  —  5  =  x2  —  2  x  +  1. 

3.  .-.  2  a;  =  6,  or  x  =  3. 

Check  :  Does  3  -  1  +  V32  -  5  =  0  ?   Does  2  +  V4  =  0  ?   No.    (§  149.) 

Therefore  3  is  not  a  root  of  the  equation.  Kecall  that  in  solving  an 
equation  a  number  is  sought  which  will  satisfy  the  equation.  The  equa- 
tion may,  however,  impose  an  impossible  relation  upon  some  numbers,  as 
in  this  case,  and  then  it  is  impossible  to  find  a  solution. 

What  is  the  explanation  of  the  solution  x  —  3  ?  If  the  original  equa- 
tion is  compared  with  the  equation  of  Example  1 ,  it  is  noticed  that  the 
only  difference  is  in  the  sign  of  the  radical ;  also  tbat  in  step  2,  after 
squaring  both  members  in  both  examples,  the  resulting  equation  is  the 
same.  In  each  example,  if  the  equation  of  step  1  has  a  root,  that  number 
is  a  root  of  the  equation  of  step  2 ;  but,  since  the  equation  of  step  2  is  the 
same  in  each  solution,  it  cannot  be  asserted  in  advance  whether  its  root 
or  roots  are  roots  of  the  equation  of  Example  1  or  of  Example  2.  When 
finally  the  solution  x  =  3  is  obtained,  the  question  arises,  is  3  a  root  of 
the  equation  in  Example  1  or  m  Example  2  ?    The  root  x  =  3  satisfies  the 


168  ALGEBRA 

equation  of  Example  1 ;  it  does  not  satisfy  the  equation  of  Example  2 
It  is  customary  to  say  that,  in  Example  2,  the  extraneous  root  3  is  intro 
duced  by  the  method  of  solution. 

This  example  makes  clear  the  necessity  of  checking  the  solutions  oi 
equations. 


Example  3.     Solve  the  equation  V#  —  2  +  V2  x  -f-  5  =  3. 


Solution  :  1.    Transposing,  vx  —  2=  3—  V2  x  + 


2.  Squaring,  x  —  2  =  9  —  6V2  x  +  5  +  2  x  +  5. 

3.  .-.  6V'2x  +  o  =  x+  16. 

4.  Squaring,  36(2  x  +  5)  =  x2  +  32  x  +  256. 

5.  .  \  a;2  -  40  z  +  76  =  0.     .-.  a  =  2,or38.     (§110.) 


Check:  Does   v7^  -  2  +  V2  .  2  +  5  =  3  ?    Does  Vo  +  V9  =  3  ?     Yes. 
Does  V38-2  +  V2  .  38  +  5  =  3  ?    Does  V36  +  V81  =  3  ?    No.     (See 
§  149.) 

Therefore  x  =  2  is  the  only  solution  of  this  equation. 

Note  1.    Jt  will  be  found  that  the  extraneous  root  38  will  satisfy  the  equa- 
tion V2z  +  5—  Vcc  -2  =  3. 

Note  2.    When  there  are  two  radicals  in  an  equation,  arrange  the  terms  so 
that  one  radical  appears  alone  in  one  member  of  the  equation. 

EXERCISE  84 
Solve  and  check  the  following  equations : 
1.    V3a-5-2  =  0.  9     V^6-fV"^-— - 


Vz-6 


2.  V6z  +  9  +  8  =  5.  

, 1A     V3r  +  1+V3r      A 

3.  V9x*  +  5-3x  =  l.  10-       ,-— —         ==4" 

4.  Vy-V2/-12  =  2.  /— —   ,     , 

5.  V^  +  4  +  V^=3.  '    VaT^a-VaT^a 

6.  \/8  x3  -  12  a,*2  +  1  =  2  a.  V2z-3_  V4a7^4 

7.  Vs  +  ll  +  V^Tor=5.  V3a;  +  2      V6a  +  1 

,—        . 2  13.    VlO  +  #-VlO-a;=2. 

8.  Vm+Vm  +  4  =  — •  

Vm  14.    V6-hl0a-3a2=2a-3. 


RADICALS  169 


15.    Vc  +  2  +  V3c  +  4  =  2.         17.    Vs8+8a>2+16a>-l=a;+3. 


16.    Vw-l  +  V3w  +  3  =  4.       18.    Vy  +  3- V2/+8  =  - Vy. 


19.    Var2  —  V2oj +  !  =  »  —  1. 


20.    V5+a;+V5  —  x  — 


12 


21. 


Vo-aj 


V*  +  2         V*         6 

Solve  for  a? : 

9a2-62 


22.    VsB  —  12  ct& 


v« 


23.    Va  +  a?  —  Va  —  #  =  Va?. 


24.  ■s/(x-2b)(x  +  Sb)=x  +  4:b. 

25.  V3a+2a—  V4 a?  —  6  a  =  V2a. 

26.  Solve  the  equation  t  =  «A/-: 
(a)   forZ;         (6)  for  gr. 

27.  Solve  the  equation  V=V^gs: 
(a)  for  #  ;         (b)  for  s. 

28.  Va  +  ic  — V^»  = 


29.    V#-a  +  V2a+3a=  V5  a. 


30.    V(a  +  26)a  —  2a&  =  a  —  4  7>. 
31-    \^T5  +  ^  +  4-2 


XV.    LOGARITHMS 


151.    Logarithms  are  exponents. 

Every  positive  number  may  be  expressed,  exactly  or  approxi- 
mately, as  a  power  of  10.  The  exponent  corresponding  to  a 
number  so  expressed  is  called  its  Logarithm  to  the  Base  10. 

Thus,  102  =  100 ;  therefore  2  is  the  logarithm  of  100  to  the  base  10, 

This  is  written  :  logi0100  =  2,  or  more  briefly  log  100  =  2. 

Similarly  log1035  is  read  "  logarithm  of  35  to  the  base  10." 


152.  Much  difficult  computation  may  be  simplified  by  the 
use  of  logarithms.  To  make  this  fact  clear,  the  approximate 
values  of  some  powers  of  10  will  be  computed  and  some  ex- 
amples will  be  solved. 

10°=1;  lO^lO;  102=100;  103=1000, 


1. 


3. 


10-*=  10'  =  V10  =  3.1623. 

101J5  =  101  x  10 5  =  10  x  3.1623  =  31.623. 

1025  =  101xlOL5=10  x  31.623=316.23. 

10-*=  (105)*  =  V33623  =  1.7782. 

101-*5  =  101  X 10  25=  10  x  1.7782  =  17.782. 

10**  =  101  x  10125=  10  X  17.782=177.82. 

1 07S=  (101-5)*  =  V3T623  =  5.6234. 
10175=  10  x  10-75=10  x  5.6234  =  56.234. 
102-75  =  10  x  10175=10  X  56.234=562.34. 

170 


1.0000  =  10000 

1.7782  =  10025 

3.1623  =  10050 

5.6234  =  10°  -7B 

10.0000  =  10100 

17.7820  =  101-25 

31.6230  =  10150 

56.2340  =  10175 

100.0000  =  10200 

177.8200  =  10225 

316.2300  =  102^° 

562.3400  =  102-75 

1000.0000  =  10300 


LOGARITHMS 


171 


Example  1.     Find  3.1623  x  17.782. 


Solution:  1.   3.1623  x  17.782 

2.  =  10-50  x  101-25  =  101-7*. 

3.  .-.  3.1623  x  17.782  =  56.234. 
This  is  approximately  correct. 


Example  2.     Find  1000  ■*•  56.234. 


Solution  :  1.    1000  -r-  56.234 

2.  =  103  -f-  10L75  as  103-1-75  =  101^5. 

3.  .-.  1000 -*- 56.234  =  17.782. 
The  solution  is  correct. 


Check  : 


Check: 


3.1623 

17.782 
63246 
252984 
221361 
221361 
31623 
66.232+ 


17.78 


56.234)1000.00000 
562  34 
437  660 
393  638 
44  0220 
39  3638 
4  65820 
4  49872 


Example  3.    Find  (5.6234)2  x  316.23  -s- 177.82. 

Solution  :    1.    (5.6234)2  x  316.23  -=- 177.82 

2.  =  (10-75) 2  X  102.50^.102.25 

3.  =  101.50+2.50-2.25  —  101-75. 

4.  .-.   (5.6234)2  x  316.23  ~  177.82  =  56.234. 

This  example  also  may  be  checked  by  ordinary  computation. 


153.  From  the  examples  of  §  152  it  is  clear  that  a  more 
complete  list  of  exponents  (logarithms)  and  ability  to  use 
them  must  be  of  great  advantage,  for  in  each  case  the  solution 
by  exponents  is  the  simpler.  The  following  paragraphs  teach 
the  methods  of  using  logarithms. 

154.  Logarithms  of  numbers  to  the  base  10  are  called 
Common  Logarithms,  and  form,  collectively,  the  Common  System 
of  Logarithms. 


172  ALGEBRA 

155.  If  a  number  is  not  an  exact  power  of  10,  its  logarithm 
can  be  given  only  approximately ;  a  four-place  logarithm  is 
one  given  correct  to  four  decimal  places. 

Thus  the  logarithm  of  13  is  1.1139  ;  i.e.  13  =  10M»,  approximately. 

The  integral  part  of  the  logarithm  is  called  the  Characteristic 
and  the  decimal  part,  the  Mantissa. 

The  characteristic  of  log  13  is  1  and  the  mantissa  is  .1139. 

Note  1.    The  plural  of  mantissa  is  mantissx. 

Note  2.    A  negative  number  does  not  have  a  logarithm. 


156.  The  Characteristic  of  the  Logarithm  of  a  Number  Greater 
than  1.     It  is  known  that  3.53  =  10-5478,  or  log 3.53  =  .5478. 

Multiplying  both  members  of  3.53  =  105478  by  10, 

35.3  =  105478  x  101  =  101-5478,  or  log  35.3  =  1.5478. 

Similarly,  353  =  101  x  1015478  =  102-5478,  or  log  353  =  2.5478. 

The  numbers  3.53,  35.3,  and  353  have  the  same  significant 
figures;  they  differ  only  in  the  location  of  the  decimal  point. 
Their  logarithms  differ  only  in  the  characteristics.  These  two 
facts  indicate  a  connection  between  the  location  of  the  decimal 
point  and  the  characteristic. 

3.53  has  one  figure  to  the  left  of  the  decimal  point;  its  logarithm  has 
as  characteristic  1  less  than  1,  or  0. 

35.3  has  two  figures  to  the  left  of  the  decimal  point ;  its  logarithm  has 
as  characteristic  1  less  than  2,  or  1. 

353  has  three  figures  to  the  left  of  the  decimal  point ;  its  logarithm  has 
as  characteristic  1  less  than  3,  or  2. 

Rule.  —  The  characteristic  of  the  common  logarithm  of  a  number 
greater  than  1  is  one  less  than  the  number  of  significant  figures  to 
the  left  of  the  decimal  point. 

Thus,  the  characteristic  of  log  357.83  is  2  ;  of  log  70390.5  is  4. 


LOGARITHMS  173 

EXERCISE  85 
What  are  the  characteristics  of  the  logarithms  of : 

1.  365.  4.   7.  7.   6.35.  10.   300506.7. 

2.  2000.         5.   16.1.  8.   60907.03.     11.   300.506. 

3.  50698.    6.  123.05.     9.  500.005.   12.  1000000. 

Tell  the  number  of  significant  figures  preceding  the  decimal 
point  when  the  characteristic  of  the  logarithm  is : 

13.    4.  14.    2.         15.    0.  16.    1.         17.    3.  18.    5. 

157.  The  Characteristic  of  the  Logarithm  of  a  Number  less  than  1. 
Dividing  both  members  of  3.53  =  105478  (§  156)  by  10, 

.353  =  105478  -- 101  =  10-5478"1.     .-.  log  .353  =  .5478  -  1. 

Dividing  both  members  of  .353  =  10,5478_1,  by  10, 
.0353  =  10-5478"1  -- 101  =  10-5478"2.     .-.  log  .0353  =  .5478  -  2. 

Similarly,  .00353  =  lO-5478"3.     .-.  log  .00353  =  .5478  -  3. 

Between  the  decimal  point  and  the  first  significant  figure  of : 

.353  there  are  no  zeros ;  the  characteristic  of  log  .353  is  —  1. 

.0353  there  is  one  zero  ;  the  characteristic  of  log  .0353  is  —2. 

.00353  there  are  two  zeros ;  the  characteristic  of  log  .00353  is 
-3. 

Rule.  —  The  characteristic  of  the  common  logarithm  of  a  number 
less  than  1  is  negative ;  numerically  it  is  one  more  than  the  number 
of  zeros  between  the  decimal  point  and  the  first  significant  figure. 

Thus,  the  characteristic  of  log  .0045  is  -  3  ;  of  log  .00027,  is  -  4. 

EXERCISE  86 
What  are  the  characteristics  of  the  logarithms  of : 

1.  .05.  3.    .00064.       •     5.    .00007.  7.    .3. 

2.  .0032.         4.   .0586.  6.   .08375.  8.   .33759. 

Tell  the  number  of  zeros  preceding  the  first  significant  figure 
when  the  characteristic  of  the  logarithm  is : 

9.    -3.         10.    -1.         11.    —5.         12.    —2.        13.    -4. 


174  ALGEBRA 

158.  Method  of  Writing  a  Negative  Characteristic.  In  §  157 
log  .353  =  .5478  -  1.  Actually,  therefore,  log  .353  is  -  .4522,  a 
negative  number.  For  many  reasons,  however,  the  positive 
mantissa,  and  the  negative  characteristic  are  retained. 

.5478  —  1  is  written :  9.5478  —  10.  Numerically  the  two  ex- 
pressions have  equal  value.     Note  that  9  —  10  =  —  1. 

The  process  in  general  is  to  decide  upon  the  characteristic 
by  the  rule  in  §  157 ;  then,  if  it  is  —  1,  write  it  9  —  10 ;  if  —  2, 
write  it  8  -<- 10 ;  etc. 

Thus,  log  .02  is  3010  -  2,  or  8.3010  -  10. 

Note.  The  negative  characteristic  is  often  written  thus:  log  .02=2.3010; 
again,  log  .353  =  1.5478.  The  minus  sign  is  written  over  the  characteristic  to 
indicate  that  it  alone  is  negative,  the  mantissa  heiug  positive. 

EXERCISE  87 

1-12.  Tell  how  each  of  the  characteristics  of  the  examples 
of  Exercise  86  should  be  written. 

159.  Mantissa  of  a  Logarithm.     From  §§  156  and  157 : 

log  3.53  =  .5478 ;  log  .353  =  9.5478  -  10 ; 
log  35.3  =  1.5478;  log  .00353  =  7.5478  -  10. 

The  numbers  3.53,  35.3,  .353,  and  .00353  have  the  same 
significant  figures.  Their  common  logarithms  have  the  same 
mantissas.     This  is  an  example  of  the 

Rule.  —  The  common  logarithms  of  all  numbers  having  the 
same  significant  figures  have  the  same  mantissae. 

Thus,  the  logarithms  of  2506,  2.506,  250.6,  etc.,  all  have  the  same 
mantissae. 

160.  A  Table  of  Logarithms  consists  of  the  mantissas  of  the 
logarithms  of  certain  numbers.  The  characteristics  of  the 
logarithms  may  be  determined  by  the  rules  given  in  §§  156 
and  157.  The  table  given  on  pages  176  and  177  gives  the 
mantissae  of  all  integers  from  100  to  999  inclusive,  calculated 


LOGARITHMS  175 

to  four  decimal  places.  The  decimal  point  is  omitted.  Such 
a  table  is  called  a  four-place  table.  While  a  five  or  six  place 
table  would  be  more  accurate,  this  table  is  sufficiently  ac- 
curate for  all  ordinary  purposes. 

161.  To  find  the  Logarithm  of  a  Number  of  Three  Significant 
Figures. 

Example  1.     Find  the  logarithm  of  16.8. 

Solution  :  1.  In  the  column  headed  "No."  find  16.  On  the  hori- 
zontal line  opposite  16,  pass  over  to  the  column  headed  by  the  figure  8. 
The  mantissa  .2253  found  there,  is  the  required  mantissa. 

2.  The  characteristic  is  1,  by  the  rule  in  §  156. 

3.  .-.log  16.8  is  1.2253. 

Rule.  —  To  find  the  logarithm  of  a  number  of  three  figures : 

1.  Look  in  the  column  headed  "  No."  for  the  first  two  figures  of 
the  given  number.  The  mantissa  will  be  found  on  the  horizontal 
line  opposite  these  two  figures  and  in  the  column  headed  by  the 
third  figure  of  the  given  number. 

2.  Prefix  the  characteristic  according  to  §  §  156  and  157. 

Example  2.     Find  log  .304. 

Solution  :   1.    Opposite  30  in  the  column  headed  by  4  is  the  mantissa 
.4829.     The  characteristic  is  -  1  or  9  -  10.     (§§  157  and  158.) 
2.   /.log  .304  =  9.4829 -10. 

Note.  The  logarithm  of  a  number  of  one  or  two  significant  figures  may 
be  found  by  using  the  column  headed  0.  Thus  the  mantissa  of  log  8.3  is  the 
same  as  the  mantissa  of  log  8.30 ;  of  log  9,  the  same  as  of  log  900. 

EXERCISE  88 
Find  the  logarithms  of : 

1.  235.  5.   72.  9.   56.2.  13.  .00465. 

2.  769.  6.   8.  10.   7.83.  14.  8690. 

3.  843.  7.  3.2.  11.    .924.  15.  24700. 

4.  900.  8.    620.  12.    .0326.  16.  60.7. 


176 


ALGEBRA 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

IO 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

o374 

ii 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

o755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1 106 

13 

"39 

"73 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

143° 

14 

1461 

1492 

1523 

x553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1 761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3U9 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

354i 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

37" 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

415° 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

43*4 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5°79 

5092 

5105 

5"9 

5132 

5M5 

5*59 

5r72 

33 

5185 

5198 

521 1 

5224 

5237 

5250 

5263 

5270 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

54i6 

5428 

35 

544i 

5453 

5465 

5478 

549o 

5502 

55H 

5527 

5539 

555i 

36 

5§o3 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

59" 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

4i 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

65U 

6522 

45 

6532 

6542 

655i 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

691 1 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7^67 

5i 

7076 

7084 

7°93 

7101 

7110 

7118 

7126 

7135 

7*43 

7152 

52 

7160 

7168 

7177 

7185 

7*93 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

73o8 

73i6 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

No. 

0 

1 

2   3 

4 

5 

6 

7 

8 

9 

LOGARITHMS 


177 


No. 

0 

1 

2 

3  |  4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

745i 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

75*3 

7520 

7528 

7536 

7543 

755i 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

773i 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

793i 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8i95 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

83S8 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8oSP 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9i54 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

93°9 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

935° 

9355 

9360 

9365 

937° 

9375 

9380 

9385 

939o 

87 

9395 

9400 

9405 

9410 

94i5 

9420 

9425 

943° 

9435 

9440 

88 

9445 

945° 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

95°4 

95°9 

95^ 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

958i 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

97°3 

9708 

97J3 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

975° 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

993o 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

178  ALGEBRA 

162.  To  find  the  Logarithm  of  a  Number  of  More  than  Three 
Significant  Figures. 

Example  1.     Find  log  327.5. 

Solution:   1.   From  the  table  :    log 327     =2.51451     _,_ 

log  328     =  2.5159J  * w* 

2.  Since  327.5  is  between  327  and  328,  its  logarithm  must  be  between 
their  logarithms.  An  increase  of  one  unit  in  the  number  (from  327  to 
328)  produces  an  increase  of  .0014  in  the  mantissa.  It  is  assumed  there- 
fore that  an  increase  of  .5  in  the  number  (from  327  to  327.5)  produces  an 
increase  of  .5  of  .0014,  or  of  .0007,  in  the  mantissa. 

3.  .-.  log  327.5  =  2.5145  +  .5  x  .0014 

.    =2.5145 +  .0007  =2.5152. 

This  result  is  obtained  in  practice  as  follows.  The  difference  between 
any  mantissa  and  the  next  higher  mantissa  as  written  in  the  table  (neglect- 
ing the  decimal  point)  is  called  the  tabular  difference.  The  tabular  dif- 
ference for  this  example  is  14(5159-5145).  .5  of  the  tabular  difference  is 
7.     Adding  this  to  5145  gives  5152,  the  required  mantissa  of  log  327.5. 

Similarly  to  find  log  327.25,  the  tabular  difference  is  14.  .25  x  14=3.5. 
Hence  the  mantissa  of  log  327.25  is  5145  +  3.5  or  5148.5.  .-.  log  327.25 
=  2.5149. 

Note  1.  The  process  of  determing  a  mantissa  which  is  between  two 
mantissae  of  the  table  is  called  Interpolation. 

Note  2.  The  assumption  made  in  step  2  is  not  warranted  by  the  facts. 
Nevertheless,  for  ordinary  purposes,  the  results  obtained  in  this  manner  are 
sufficiently  correct.    This  is  the  common  method  of  interpolating. 

Note  3.  When  interpolating,  it  is  customary  to  cut  down  all  decimals  so 
that  the  mantissa  will  again  be  a  four-place  decimal.  Thus  3.5  is  called  4. 
3.4  would  be  called  3. 

Rule.  —  To  find  the  logarithm  of  a  number  of  more  than  three 
significant  figures : 

1.  Find  the  mantissa  for  the  first  three  figures,  and  the  tabular 
difference  for  that  mantissa. 

2.  Multiply  the  tabular  difference  by  the  remaining  figures  of 
the  given  number,  preceded  by  a  decimal  point. 

3.  Add  the  result  of  step  2  to  the  mantissa  obtained  in  step  1. 

4.  Prefix  the  proper  characteristic  by  §§  156  and  157. 


LOGARITHMS 


179 


Example  2.     Find  log  34.652. 
Solution  :  1.    Mantissa  of  log  346  =  5391. 
Mantissa  of  log  347  =  5403. 

2.  Tabular  difference  =  12.     .52  x  12  =  6.24  =  6. 

3.  .-.  Mantissa  for  log  34652  =  5391  +  6  =  5397. 

4.  .-.  log  34.652  =  1.5397. 

Example  3.     Find  log  .021508. 
Solution  :  1.     Mantissa  of  log  215  =  3324. 
Mantissa  of  log  216  =  3345. 

2.  Tabular  difference  =±  21.     .08  x  21  =  1.68  =  2. 

3.  .  •.  mantissa  of  log  21508  =  3324  +  2  =  3326. 

4.  .-.  log  .021508  =  .3326  -  2,  or  8.3326  -  10. 

EXERCISE  89 
Find  the  tabular  difference  for  the  mantissae: 

1.  3222.         3.   6590.        5.   8982.        7.   7076. 

2.  4166.         4.   7364.        6.   5340.        8.   8692. 

Find  the  logarithms  of : 


11.  325.5. 

12.  263.1. 

13.  786.3. 

14.  492.2. 

15.  703.4. 


16.  32.16. 

17.  1.608. 

18.  7.961. 

19.  .8462. 

20.  .05375. 


21.  327.11. 

22.  243.25. 

23.  62.721. 

24.  803.75. 

25.  6.2534. 


9.   4728. 
10.   7435. 

26.  3.1416. 

27.  1.0453. 

28.  .22735. 

29.  .063457. 

30.  .004062. 


163.    To  find  the  Number  Corresponding  to  a  Given  Logarithm. 

Example  1.     Find  the  number  whose  logarithm  is  1.6571. 

Solution  :  1.     Find  the  mantissa  6571  in  the  table. 

2.  In  the  column  headed  "No."  on  the  line  with  6571  is  45.  These 
are  the  first  two  figures  of  the  number.  At  the  head  of  the  column  con- 
taining 6571  is  4,  the  third  figure  of  the  number.  Hence  the  number 
sought  has  the  figures  454. 


180  ALGEBRA 

3.   The  characteristic  being  1,  the  number  must  have  two  figures  to 
the  left  of  the  decimal  point.      (§  156.) 
.-.  the  number  is  45.4. 

Rule.  —  To  find  the  number  corresponding  to  a  given  logarithm 
when  the  mantissa  appears  in  the  table : 

1.  Find  the  three  figures  corresponding  to  this  mantissa,  as  in 
the  example. 

2.  Place  the  decimal  point  according  to  the  rules  in  §  §  156  and 
157. 

EXERCISE  90 

Find  the  numbers  whose  logarithms  are : 

1.  2.6138.         4.   2.9542.         7.   1.7404.         10.   9.8000-10, 

2.  1.3365.         5.    3.9289.         8.   4.7024.         11.    8.5378-10. 

3.  3.6972.         6.   0.8162.         9.   0.8893.         12.    7.4133-10. 

Example  2.     Find  the  number  whose  logarithm  is  1.3934. 

Solution  :  1.   The  mantissa  3934  does  not  appear  in  the  table. 

The  next  less  mantissa  is  3927,  and  the  next  greater  is  3945. 

The  correspoHding  numbers  are  247  and  248.  That  is  : 
mantissa  of  log  247  =  3927  1  Diff.  ]  Tabular 
mantissa  of  log  x  =  3934  J  =  7.  i  difference 
mantissa  of  log  248  =  3945  j  =  18. 

2.  Since  an  increase  of  18  in  the  mantissa  produces  an  increase  of  1 
in  the  number,  it  is  assumed  that  an  increase  of  7  in  the  mantissa  must 
produce  an  increase  of  ^g  or  .38  in  the  number.  Hence  the  number  has 
the  figures  247.38. 

3.  Since  the  characteristic  is  1,  the  number  must  be  24.738. 

Rule.  —  To  find  the  number  corresponding  to  a  given  logarithm 
when  the  mantissa  does  not  appear  in  the  table : 

1.  Find  in  the  table  the  next  less  mantissa.  Find  the  correspond- 
ing number  of  three  figures,  and  the  tabular  difference. 

2.  Subtract  the  next  less  mantissa  from  the  given  mantissa  and 
divide  the  remainder  by  the  tabular  difference. 


LOGARITHMS  181 

3.  Annex  the  quotient  to  the  number  of  three  figures  obtained  in 
step  1. 

4.  Place  the  decimal  point  according  to  the  rules  in  §§  156  and 
157. 

EXERCISE  91 
Find  the  numbers  whose  logarithms  are  : 

1.  1.8079.  6.   0.8744.  11.  2.5369. 

2.  3.3565.  7.    9.9108-10.  12.  9.7022-10. 

3.  2.6639.  8.   8.8077-10.  13.  2.4644. 

4.  0.7043.  9.    7.5862-10.  14.  3.1634. 

5.  2.5524.  10.   8.2998-10.  15.  2.9310. 

PROPERTIES  OF  LOGARITHMS 

164.  The  preceding  discussion  relates  entirely  to  the  Com- 
mon System  of  Logarithms.  (§  154.)  Certain  properties  of 
logarithms  to  any  base  will  be  considered  now. 

Note.    The  base  may  be  any  positive  number  different  from  1. 

165.  Just  as  log103.053  =  .4847  means  that  10-4847  =  3.053, 
so  \oga]¥—x  means  that  J¥=ax. 

Loga  N  is  read  "  the  logarithm  of  N  to  the  base  a." 


166.    Logarithm  of  a  Product 

Assume  that  ax  = 
and  ay 


X==^l;then   (*  =  j°&.^ 


Also  a*  -a"=MN,  or  ax+v=MN.     .\\ogaMN=x+y.  (§165) 
Therefore  loga  MN=  loga  M  +  loga  N. 

Rule.  —  In  any  system,  the  logarithm  of  a  product  is  equal  to  the 
sum  of  the  logarithms  of  its  factors. 

Example  1.     Given   log  2  =  .3010,  and   log  3  =  .4771,   find 
log  72. 


182 


ALGEBRA 


Solution  :  1.   log  72  =  log  2  •  2  .  2  .  3  .  3. 

2.  =  log2  +  log  2  + log  2  + log  3  +  log3. 

3.  .-.  Iog72  =  31og2  +  2  1og3  =  3(.3010)  +2 (.4771) 

=  .9030 +  .9542  =  1.8572. 


EXERCISE  92 

Given  log  2  =  .3010,  log  3  =  .4771,  and  log  7  =  .8451.  Find 
the  following  logarithms  as  in  Example  1 ;  check  the  solutions 
by  finding  the  same  logarithms  in  the  table : 

1.  log  21.  4.    log  126.  7.    log  324. 

2.  log  42.  5.    log  128.  8.    log  378. 

3.  log  36.  6.    log  252.  9.   log  168. 
10.  Find  by  logarithms  the  value  of  35.2  X  2.35  x  6.43. 


Solution  :  1.   Let  v  =  35.2  x  2.35  x  6.43. 

2.  .  \  log  v  =  log  35.2  +  log  2.35  +  log  6.43. 

3.  .-.  log  v-  2.7258. 

4.  .-.       t>  =  531.87.    (§163.) 


log  35.2 
log  2.35 
log  6.43 


1.5465 
0.3711 

0.8082 
2.7258 


Find  by  logarithms  the  values  of : 

11.  32.5x27.8.  14.   34.55x29.9.      17.   3.142x6039. 

12.  2.49x65.7.  15.    678.1x37.        18.   541.2x1.523. 

13.  .289  x  365.  16.    1.732  x  580.       19.   43.65  x  865.25. 
20.   Find  by  logarithms  the  value  of  .0631  X  7.208  x  .51272. 

Solution  :  1.  log  a  =  log  .0631  +  log  7.208  +  log  .51272. 

log.  0631  =    8.8000-10 
log  7. 208=    0.8578 
log  .51272  =    9.7099 


3.  /.log  0  =  9.3677 -10. 


19.3677  -  20  =  9.3677  -  10 
,2332.      (§  163.) 


Note.    If  the  sum  of  the  logarithms  is  a  negative  number,  the  result 
should  be  written  so  that  the  negative  part  of  the  characteristic  is  — 10. 


LOGARITHMS  183 

Eind  by  logarithms  the  values  of: 

21.  .0235x3.14.  24.  84.75  x  .00368. 

22.  .5638  x  .0245.  25.  .0273  x  .00569  x  .684. 

23.  .7783x6.282.  26.  .2908  X  .0305  x  .0062. 


167.   Logarithm  of  a  Quotient. 

Assume  that       ax  =  M\ 
and  o*  =  N 


;then      1*  =  }°*- 


x  =  loga  3f, 


Also,  ax  -7-  ay  =  M -f-  JVJ  or  ax-y  =  M-^K 

.'.  \o%a(M+N)  =  x  —  y. 
Therefore,  loga  (Jf  -~  N)  =  \ogaM—  loga  iV. 

Rule.  —  In  any  system,  the  logarithm  of  the  quotient  of  two  num- 
bers is  equal  to  the  logarithm  of  the  dividend  minus  the  logarithm 
of  the  divisor. 

Example  1.     Given   log 2  =  .3010   and   log 3  =.4771,   find 

Solution  :  1.   log  f  =  log  3  -  log  2  =  .4771  -  .3010  =  .1761. 

Example  2.     Eind  log  -J. 

2  •  2  •  2 

Solution  :  1.    log  f  =  log • 

3  •  3 

2.  =(log2  +log2  +  log  2) -(log  3  +  log  3). 

3.  =  3  (.3010)-  2  (.4771)  =  .9030  -  .9542. 

4.  .-.  log  f  =  9.9488-  10. 


.9030  =  10.9030  -  10 
.9542  =      .9542 

9.9488  -  10 


Note  1.  To  find  the  logarithm  of  a  fraction,  add  the  logarithms  of  the 
factors  of  the  numerator,  and  from  the  result  subtract  the  sum  of  the  loga- 
rithms of  the  factors  of  the  denominator. 

Note  2.  To  subtract  a  greater  logarithm  from  a  less,  or  to  subtract  a 
negative  logarithm  from  a  positive,  increase  the  characteristic  of  the  minuend 
by  10,  writing — 10  after  the  mantissa  to  compensate.  Thus,  in  this  example, 
.9542  is  greater  than  .9030 ;  therefore,  .9030  is  written  10.9030  —  10,  after  which 
the  subtraction  is  performed. 


184  ALGEBRA 

EXERCISE  93 
Given  log  2  =.3010,  log  3  =  .4771,  and  log  7  =  .8451,  find: 

1.  logf  3.    logf  5.    log  |f  7.    logf 

2.  log-V-.  4.    log ty.  6.    log-V-.  8.    logji 

Find  by  logarithms  the  values  of: 

9.   255-48.  12.   630.5-402.  15.  2865-1.045. 

10.  376-83.  13.   300.25-3.14.  16.  7.835-23.75. 

11.  299-99.  14.   230.56-1.06.  17.  9.462-85.64. 

3.14  x  25  2     .0036  x  2.35 

365  .0084 

23.5  x  1.05  2      287.5  x  .096 

3785       '  '         3.1416 

24.75  x  .0058  25.6  x  .738  x  .0535 

1.41          '  '            265x432 

16.08  x  256  1.405  x  207  x  .00392 

17         '  "            508  x. 6354 

168.  The  Logarithm  of  a  Power  of  a  Number. 

Assume  that  ax  =  M;  then  x  =  logd  M. 

Also,      (ax)p  =  Mp,  or  apx  =  Mp.     .-.  log  Mp  =  px. 
Therefore,  log  Mp  =p  loga  M. 

Rule.  —  In  any  system,  the  logarithm  of  any  power  of  a  number 
is  equal  to  the  logarithm  of  the  number  multiplied  by  the  exponent 
indicating  the  power. 

Example  1.     Given  log  7  =  .8451,  find  log  75. 
Solution  :  log  75  =  5  log  7  =  5  x  .8451  =  4.2255. 

Example  2.     Find  by  logarithms  1.0410. 

Solution  :  1.    log  1.04i°  ==  10  log  1.04  =  10  x  .0170  =  .1700. 

2.  The  number  whose  logarithm  is  .1700  is  1.479.  (§  163) 

3.  .-.    1.0413  =  1.479. 


LOGARITHMS  185 

Example  3.     Find  by  logarithms  V365. 
Solution  :  1.    log  ^365  =  log  365?  =  |  log  365. 

2.  /.  log  ^365  =  i  x  2.5623  =  0.8541. 

3.  The  number  whose  logarithm  is  0.8541  is  7.146.  (§  163) 

4.  /.   ^365=7.146. 

When  finding  a  cube  root,  the  logarithm  of  the  radicand  is 
divided  by  3 ;  when  finding  a  square  root,  the  logarithm  of  the 
radicand  is  divided  by  2.     This  suggests  the 

Rule.  —  In  any  system,  the  logarithm  of  a  root  of  a  number  is 
the  logarithm  of  the  radicand  divided  by  the  index  of  the  root. 

Example  4.     Find  by  logarithms  a^-0359. 
Solution  :  1.     log  \Z.0359  =  \  log  .0359  =  \  (8.5551  -  10). 

2.  /.  log  -t/AJSW  =  \  (38.5551  -  40).     (See  note.) 

3.  .-.  log  \/.0359=  9.6387  -10. 

4.  The  number  whose  logarithm  is  9.6387  -  10  is  .4352.  (§  163) 

5.  /.    v'^0359  =  .4352. 

Note.  To  divide  a  negative  logarithm,  write  it  in  such  form  that  the 
negative  part  of  the  characteristic  may  be  divided  exactly  by  the  divisor,  and 
give  —  10  as  quotient. 

Thus  8.5551  —  10  is  changed  to  38.5551—40  since  the  divisor  is  4.  If  the 
divisor  were  3,  it  would  be  changed  to  28.5551  —  30. 

EXERCISE  94 
Given  log  2  =  .3010,  log  3  =  .4771,  and  log  7  =  .8451 ;  find : 

1.  log  37.         3.    log  74.         5.    log  (21)i         7.    log  \/6. 

2.  log  25.         4.   log  273.       6.   log  a/7.  8.    log  a^H. 

Find  by  logarithms  the  values  of  the  following: 
9.    2352.  13.    3.1416  x  182.  17.    V^ftp. 


10.  2.0453.        14.    7.7954.  18.    V25  x  19.6  x  17.3. 

11.  -y/KSE.       15.    122.  19.    V3  x  a/5. 

12.  a/97863.    16.   |  x  3.1416  x  53.     20.    (UwrY- 


186  ALGEBRA 

21.  The  volume  of  a  right  circular  cylinder  is  given  by  the 
formula  V=irB2H. 

Find  the  volume  (by  logarithms); 

(a)  if  P  =  10.5  and  H=  26.5. 

(b)  if  i2  =  8.2  and  #  =  33.1. 

22.  The  volume  of  a  sphere  is  given  by  the  formula  V=  f  irR3. 
Find  the  volume : 

(a)   ifP  =  12;     (b)   HE  =  6.2. 

23.  The  interest  on  P  dollars  at  r  %  for  t  years  is  given  by 

the  formula  /=  — .     Find  7: 
100 

(a)   if  P=  $  765,  r  =  5,  and  t  =  6.5  years. 

(6)   if  P  =  $  1250,  r  =  4.5,  and  £  =  8  years  and  3  months. 

24.  The  amount  to  which  P  dollars  will  accumulate  at  r% 
compound  interest  in  n  years  is  given  by  the  formula, 

(a)   if  P=  $ 250,  r  =  4,  and  n  =  10. 
(6)   if  P  =  $  75,  r  =  3.5,  and  n =  15. 

25.  A  cylindrical  cistern  has  for  its  diameter  5  feet.  Find 
the  number  of  barrels  of  water  this  cistern  has  in  it  when  the 
water  is  9  feet  deep.  (One  cubic  foot  of  water  is  about  1\ 
gallons ;  one  barrel  contains  31^-  gallons.) 

Historical  Note.  Logarithms  were  introduced  by  John  Napier  (1550- 
1617),  a  Scotch  gentleman  who  studied  mathematics  and  science  as  a  pastime. 
The  Napier  logarithms  were  not  the  common  logarithms.  Briggs  (1556-1631) , 
an  English  mathematician,  computed  the  first  table  of  Common  Logarithms. 


XVI.   PROGRESSIONS 

ARITHMETIC  PROGRESSION 

169.  An  Arithmetic  Progression  (A.  P.)  is  a  sequence  of 
numbers,  called  terms,  each  of  which  after  the  first  is  derived 
from  the  preceding  by  adding  to  it  a  fixed  number,  called  the 
Common  Difference. 

Thus,  1,  3,  5,  7,  •••  is  an  A.  P.  Each  term  is  derived  from  the  preced- 
ing by  adding  2.  The  next  two  terms  are  9  and  11.  2  is  the  common 
difference. 

Again,  9,  6,  3,  •••  is  an  A.  P.  The  common  difference  is  —  3.  The 
next  two  terms  are  0  and  —  3. 

Note.  The  common  difference  may  be  found  by  subtracting  any  term 
from  the  one  following  it. 

EXERCISE  95 

Determine  which  of  the  following  are  arithmetic  progres- 
sions; determine  the  common  difference  and  the  next  two 
terms  of  the  arithmetic  progressions  : 

1.  4,  7,  10,  13,  ....  6.  5  m,  7.5  m,  10m,  .... 

2.  1,  3,  7,  9,  15,  ....  7.  4p,  1.5 p,  -jp,  .... 

3.  10,  7,  4,  1,  ....  8.  1.06,  1.12,  1.18,  .... 

4.  25,20,15,10,....  9.  a  +  b,a  +  2b,  a  +  36,  .... 

5.  21,  3£,  4,  4f,  ....  10.  5r+6s,6H-4s,7r+2s,.... 
Write  the  first  five  terms  of  the  A.  P.  in  which : 


the  first  term  is 

the  common  difference  is 

11 

12 

13 

14 

15 

a 

d 

15 
6 

25 
-8 

7.5 
3.5 

X 

-4 

187 


188  ALGEBRA 

170.  The  nth  Term  of  an  Arithmetic  Progression.  It  is  possi- 
ble to  determine  a  particular  term  of  an  arithmetic  progression 
without  rinding  all  of  the  preceding  terms. 

Given  the  first  term  a,  the  difference  d,  and  the  number  of 
the  term  n,  of  an  arithmetic  progression,  find  the  nth  term  I. 

The  progression  is  a,  a-\-d,  a-\-2d,  a-\-3d,  •  •  •.  The  coeffi- 
cient of  d  in  each  term  is  1  less  than  the  number  of  the  term. 
Thus,  the  10th  term  would  be  a  +  9  d.  Therefore  the  coeffi- 
cient of  d  in  the  nth  term  must  be  (n  —  1). 

.-.  l=a+(n-l)d. 

Example.     Find  the  10th  term  of  8,  5,  2,  •••. 

Solution:    1.    a  =  8  ;  d  =-  3  ;  n  =  10  ;  I  =  ? 

2.    l  =  a+(n-l)d.     /.  Z  =  8+(10-l)(-3)=8-27=-19. 

EXERCISE  96 
Find: 

1.  The  12th  term  of  3,  9,  15,  ••• ;  also  the  20th. 

2.  The  15th  term  of  16,  12,  8,  ...  j  also  the  25th. 

3.  The  13th  term  of  -  7,  — 12,  -  17,  ••• ;  also  the  31st. 

4.  The  16th  term  of  2,  2},  3,  ••• ;  also  the  51st. 

5.  The  11th  term  of  1.05,  1.10,  1.15,  ••• ;  also  the  26th. 

6.  What  term  of  the  progression  5,  8,  11,  •••  is  86  ? 
Solution  :1.    a  =  5  ;  d  =  3  ;  I  =  86  ;  find  n. 

2.  l  =  a+(n-  l)d.     /.  86  =  5 +(n- 1)3. 

3.  Solving  for  n,  n  =  28.     .-.  86  is  the  28th  term. 

7.  What  term  of  the  progression  8,  5,  2,  •••is  —70? 

8.  What  term  of  the  progression  \,  -f,  |,  •••  is  20^? 

9.  What  term  of  the  progression  —75,  —67,  —59,  •••  is 
197? 

10.    What  term  of  the  progression  1,  1.05,  1.10,  ...  is  2  ? 


PROGRESSIONS  189 

11.  If  the  first  term  of  an  A.  P.  is  15,  and  the  11th  term  is 
35,  what  is  the  common  difference  ? 

Hint:   35  =  15  +(11  -  l)d. 

Find  the  common  difference : 

12.  If  the  first  term  is  5  and  the  22d  term  is  173. 

13.  If  the  first  term  is  —  20  and  the  33d  term  is  —4 

14.  If  the  first  term  is  325  and  the  31st  term  is  25. 

15.  Find  the  10th  term  of  the  arithmetic  progression  whose 
first  term  is  7  and  whose  16th  term  is  97. 

16.  A  man  is  paying  for  a  lot  on  the  installment  plan. 
His  payments  the  first  three  months  are  $10.00,  $10.05,  and 
$10.10.     What  will  his  20th  and  25th  payments  be  ? 

171.  The  terms  of  an  arithmetic  progression  between  any 
two  other  terms  are  called  the  Arithmetic  Means  of  those  two 
terms. 

Thus,  the  three  aiithmetic  means  of  2  and  14  are  5,  8,  11,  since  2,  5, 
8,  11,  14  form  an  arithmetic  progression. 

A  single  arithmetic  mean  of  two  numbers  is  particularly  im- 
portant.    It  is  called  The  Arithmetic  Mean  of  the  numbers. 

When  two  numbers  are  given,  any  specified  number  of 
arithmetic  means  may  be  inserted  between  them. 

Example.  Insert  five  arithmetic  means  between  13  and 
-11. 

Solution  :  1.  There  results  an  arithmetic  progression  of  7  terms,  in 
which  a  =  13,  I  =—  11,  and  n  =  7.     Find  d. 

2.  l  =  a+(n—l)d.     .*.  —  11  =  13  +  6  d,  or  d  =-  4. 

3.  The  progression  is  :    13,  9,  5,  1,  -  3,  —  7,  —  11. 

Check  :   There  is  an  A.  P.  with  five  terms  between  13  and  —  11. 


190  ALGEBRA 

EXERCISE  97 

1.  Insert  three  arithmetic  means  between  3  and  19. 

2.  Insert  four  arithmetic  means  between  —  10  and  20. 

3.  Insert  nine  arithmetic  means  between  3  and-  28. 

4.  Insert  five  arithmetic  means  between  -J  and  5. 

5.  Insert  five  arithmetic  means  between  —  j  and  —5. 

6.  Find  the  arithmetic  mean  of  7  and  15. 

7.  Find  the  arithmetic  mean  of  V2  and  Vl8. 

8.  Find  the  arithmetic  mean  of  x  -f  7  and  x  —  7. 

9.  Find  the  arithmetic  mean  of  a  and  b.  From  the  result, 
make  a  rule  for  finding  the  arithmetic  mean  of  any  two 
numbers. 

Note.  The  arithmetic  mean  of  two  numhers  is  commonly  called  their 
average. 

10.  Find  the  common  difference  if  two  arithmetic  means  are 
inserted  between  r  and  s. 

11.  Find  the  common  difference  if  k  arithmetic  means  are 
inserted  between  m  and  p. 

172.  The  Sum  of  the  First  n  Terms  of  an  Arithmetic  Pro- 
gression. 

Given  the  first  term  a,  the  nth  term  I,  and  the  number  of 
terms  n;  find  the  sum  of  the  terms  S. 

Solution:  1.   S=a+(a  +  d)  +  (a+2d)+  •••  +(l-2d)  +  (l-d)+L  (1) 

2.  Writing  the  terms  in  reverse  order, 

S  =  l+(l-d)  +  (l-2d)+...  +  (a  +  2d) +  («  +  <*)+ a.         (2) 

3.  Adding  the  equations  (1)  and  (2),  term  for  term, 

2fl  =  (a  +  Z)  +  (a  +  ?)  +  (a  +  0+  •••  +(a  +  0  +  («+0  +  («  +  0-      (3) 

4.  There  were  n  terms  in  the  right  member  of  (1)  ;  from  each,  there 
results  a  sum  (a  +  I)  in  (3). 

/.  2  S  =  n{a  +  I) ,  or  S  =  £  (a  +  /)  (4) 


PROGRESSIONS  191 

6.   In  §  288,  .  =  a  +(n  —  l)d;  substituting  this  value  of  I  in  (4), 

S  =  ^{a  +(a  +0  -  1)<Z)},  or  S  =  ?{2a  +  (n  -  l)d}.  (6) 

2     -  2 

Example  1.  Find  the  sum  of  the  first  15  terms  of  the 
arithmetic  progression,  of  which  the  first  term  is  5  and  the 
15th  term  is  45. 

Solution  :   1.   a  =  5  ;  I  =  45  ;  n  =  15. 

2.   tf  =  p(a  +  Z).     .-.#  =  ^-(5  +  45)  =  15-25  =  375. 
2 

Example  2.  Eind  the  sum  of  the  first  12  terms  of  the  pro- 
gression 8,  5,  2,  •••. 

Solution  :   1.    a  =  8  ;  d  =  —  3  ;  w  =  12. 

2.   S  =  £{2a+(»-l)d}.     ,\#  =  6{16  + 11.  (-8)}=  6(16-83}. 
2 

.-.  /S'  =  6(-17)  =  -102. 

EXERCISE  98 
Eind  the  sum  of : 

1.  12  terms  of  3,  9,  15,  •••. 

2.  15  terms  of  -  7,  -12,  -  17,  .... 

3.  16  terms  of  -  69,  -  62,  -  55,  •••. 

4.  10  terms  of  $1.06,  $1.12,  $1.18,  .... 

Eind  the  sum  of  the  terms  of  an  arithmetic  progression  if : 

5.  The  number  is  12,  the  first  is  5,  and  the  last  is  50. 

6.  The  number  is  31,  the  first  is  40,  and  the  last  is  0. 

7.  The  number  is  18,  the  first  is  — 18,  and  the  last  is  22. 

8.  The  number  is  8,  the  first  is  —  f,  and  the  last  is  ^ 

9.  Eind  the  sum  of  the  numbers  1,  2,  3,  •••,  100. 

10.  Eind  the  sum  of  the  even  numbers  from  2  to  100. 

11.  Eind  the  sum  of  the  odd  numbers  from  1  to  99. 


192  ALGEBRA 

12.  Find  the  sum  of  all  even  integers,  beginning  with  2  and 
ending  with  250. 

13.  If  a  boy  earns  $360  during  his  first  year  of  work,  and 
is  given  an  increase  of  $50  per  year  for  each  succeeding  year, 
what  is  his  salary  during  his  10th  year,  and  how  much  has  he 
earned  altogether  during  the  10  years  ? 

14.  If  at  the  beginning  of  each  of  10  years  a  man  invests 
$100  at  6%  simple  interest,  to  what  does  the  principal  and 
interest  amount  at  the  end  of  the  10th  year  ? 

15.  How  many  poles  will  there  be  in  a  pile  of  telegraph 
poles  if  there  are  25  in  the  first  layer,  24  in  the  second,  etc., 
and  1  in  the  last  ? 

16.  A  man  has  a  debt  of  $3000,  upon  which  he  is  paying 
6  °/o  interest.  At  the  end  of  each  year  he  plans  to  pay  $  300 
and  the  interest  on  the  debt  which  has  accrued  during  the 
year.  How  much  interest  will  he  have  paid  when  he  has 
freed  himself  of  the  debt  ? 

17.  A  man  is  paying  for  a  $300  piano  at  the  rate  of  $10 
per  month  with  interest  at  6%.  Each  month  he  pays  the  total 
interest  which  has  accrued  on  that  month's  payment.  How  much 
money,  including  principal  and  interest,  will  he  have  paid 
when  he  has  freed  himself  from  the  debt? 

18.  It  has  been  learned  that,  if  a  marble,  placed  in  a  groove 
on  an  inclined  plane,  passes  over  a  distance  D  in  one  second, 
then  in  the  second  second  it  will  pass  over  the  distance  3  Z),  in 
the  third,  over  the  distance  5  Z>,  etc.  Over  what  distance  will 
it  pass  in  the  10th  second  ?    in  the  tth.  second. 

19.  Through  what  total  distance  does  it  pass  in  5  seconds  ? 
in  10  seconds  ?   in  t  seconds  ? 


PROGRESSIONS  193 

20.  Experiment  has  shown  that  an  object  will  fall  during 
successive  seconds  the  following  distances : 

1st  second,  16.08  ft. ;  3d  second,  80.40  ft. ; 

2d  second,  48.32  ft. ;  4th  second,  112.56  ft. 

Find  the  distance  through  which  the  object  will  fall  during 
the  7th  second ;  the  tth  second. 

21.  Find  the  total  distance  through  which  the  object  falls  in 
5  seconds  ;  in  t  seconds. 

22.  Substitute  g  for  32.16  in  the  final  result  of  Example  21 
and  simplify  the  result. 

173.  In  an  arithmetic  progression,  there  are  five  elements, 
a,  d,  I,  n,  S.  Two  independent  formulae  connect  these  ele- 
ments, the  formula  for  the  sum  and  the  formula  for  the  term  I. 
Hence  if  any  three  of  the  elements  are  known,  the  other  two 
may  be  found. 

Note.    Remember  that  the  formula  for  the  sum  is  given  in  two  ways. 

Example  1.     Given  a  —  —  f,  w  =  20,  S  =—  f ;  find  d  and  I. 

Solution:  1.   S  =  -(a+l).     .'.--=  10  f  --+  A  ;  whence  Z_-. 
2V  '  3  V     3        )  2 

2.   l  =  a  +  (n-l)d.    .-.  f  =-  §  +  (19)  •  d\  whence  d  =  |. 


Example  2.     Given  a  =  7,  d  =  4,  S  =  403 ;  find  n  and  I. 

Solution  :  1.    S  =  -{2  a  +(»-  l)d}.     .'.  403  =  -{14  +  (n  -  1)  •  4}. 
2  2 

I.   .-.  806  =  n{4  n  +  10}  ;  4  n2  +  10  n  -  806  =  0  ;  2  n2  +  5  n  -  400  =  0. 


.    „_-5±V25  +  3224  5±v3249      _5.fc.57  62    nrt   ,   1Q 

•  .  /i  — — —  '        —  —  — — ,  or  -j-  i-o* 

4  4  4  4' 

Since  n  is  the  number  of  terms,  n  must  be  13. 

3.   I  =  a  +  (n  -  \)d.     .-.  I  =  7  +  12  •  4  =  55. 

Note.    A  negative  or  a  fractional  value  of  n  is  inapplicable,  and  must  be 
rejected  together  with  all  other  values  depending  upon  it. 


194  ALGEBRA 

Example  3.     The  sixth  term  of  an  arithmetic  progression 
is  10  and  the  16th  term  is  40.     Find  the  10th  term. 
Solution  :  1.  By  the  formula  I  =  a  4-  {n  —  l)d : 
a  +  5  d  -  10. 
a  +  15^=40. 

2.  Solving  the  system  of  equations  in  step  1,  d  —  3  and  a  =  —  6. 

3.  The  10th  term  :  Z  =  -  5  +  9  •  3  =-  5  +  27  =  22. 

EXERCISE  99 

1.  Given  d  =  5,  1=71,  n  =  15 ;  find  a  and  S. 

2.  Given  a  =  -  9,  n  =  23,  I  =  57;  find  d  and  & 

3.  Given  a  =  J,  Z  =  *£,  5  =»*$*}  find  d  and  ». 

4.  Given  a  =  \,  Z  =  —  yV>  d  =  —  -jj ;  find  .w  and  #. 

5.  Given  d  =  J,  n  =  17,  S  =  17 ;  find  a  and  Z. 

6.  Given  a  =  f,  n»15,  #=*±$*j  find  d  and  I 

7.  Given  a  =  -f,  Z  =  --2/-,  #  =  -91;  find  d  and  n. 

8.  Given  a  =  -1/,  d  =  -  § ,  5  -*  If*  •  find  n  and  & 

9.  Given  a,  I,  and  n ;  derive  a  formula  for  d. 

10.  Given  a,  d,  and  Z;  derive  a  formula  for  n. 

11.  Given  a,  n,  and  £;  derive  a-formula  for  Z. 

12.  Given  d,  n,  and  # ;  derive  a  formula  for  a. 

13.  Given  d,  I,  and  n ;  derive  formulae  for  a  and  #. 

14.  The  8th  term  of  an  arithmetic  progression  is  10,  and  the 
14th  term  is  -  14.     Find  the  23d  term. 

15.  The  7th  term  of  an  arithmetic  progression  is  —  J,  the 
16th  term  is  J,  and  the  last  term  is  -1/-.    Find  the  number  of  terms. 

16.  The  sum  of  the  2d  and  6th  terms  of  an  arithmetic  pro- 
gression is  —  j-,  and  the  sum  of  the  5th  and  9th  terms  is  —  10. 
Find  the  first  term. 

17.  Find  four  numbers  in  arithmetic  progression  such  that 
the  sum  of  the  first  two  shall  be  12,  and  the  sum  of  the  last 
two  -  20. 


PROGRESSIONS  195 

18.  Find  five  numbers  in  arithmetic  progression  such  that 
the  sum  of  the  second,  third,  and  fifth  shall  be  10,  and  the 
product  of  the  first  and  fourth  —  36. 

19.  Find  three  numbers  in  arithmetic  progression  such,  that 
the  sum  of  their  squares  is  347,  and  one  half  the  third  number 
exceeds  the  sum  of  the  first  and  second  by  4i 

20.  Find  three  integers  in  arithmetic  progression  such  that 
their  sum  shall  be  12,  and  their  product  —  260. 

GEOMETRIC  PROGRESSION 

174.  A  Geometric  Progression  (G-.  P.)  is  a  sequence  of  num- 
bers, called  terms,  each  of  which,  after  the  first,  is  derived  by 
multiplying  the  preceding  term  by  a  fixed  number  called  the 
Ratio. 

Thus,  2,  6,  18,  54,  •••  is  a  geometric  progression.  Each  term  is  ob- 
tained by  multiplying  the  preceding  term  by  3.     The  ratio  is  3. 

Again,  15,  —  5,  -f  f,  —  §,  •••  is  a  G.  P.  The  ratio  is  —  \.  The  next 
two  terms  are  +  ^7  and  —  /T. 

Note.  The  ratio  may  be  found  by  dividing  any  term  by  the  one  preced- 
ing it. 

EXERCISE  100 

Determine  which  of  the  following  are  geometric  progres- 
sions ;  determine  the  ratio  and  also  the  next  two  terms  of  the 
geometric  progressions : 

1.  4,  8,  16,  32,  •-.  6.   3x,  6x2,  12  a?,  — 

2.  200,  50,  25,  10,  ....  7.   2,  -4,  -  8,  16,  -32,  .... 

3.81,27,9,....  8'    (1+^(1  +  ^(1+^-. 

9    1      1      1 

4.  -2,  +6,  -18,  +54,  ....     y-  m>  — 2>  ms>      * 

K       5m    5m  tn    10    2      2 

5.  5m,— ,—,....  10.   _,-,_,.... 


196  ALGEBRA 

Write  the  first  five  terms  of  the  G.  P.  in  which : 


11 

12 

13 

14 

15 

The  first  term  is  : 
The  ratio  is  : 

-5 

-2 

100 

1 

3 

2 

a 
r 

175.  The  nth  Term  of  a  Geometric  Progression.  It  is  possible 
to  determine  a  particular  term  of  a  geometric  progression 
without  finding  all  of  the  preceding  terms. 

Given  the  first  term  a,  the  ratio  r,  and  the  number  of  terms 
n,  of  a  geometric  progression,  determine  the  nth  term  I 

The  progression  is  a,  ar,  ar2,  ar3,  •••. 

The  exponent  of  r  in  each  term  is  1  less  than  the  number  of 
the  term.  Hence  the  10th  term  would  be  ar9.  Therefore  the 
exponent  of  r  in  the  nth  term  must  be  (n  —  1). 

.-.  l=arn~\ 

Example.     What  is  the  7th  term  of  9,  3,  1,  •••  ? 
Solution  :  1.    a  =  9;r  =  £;n  =  7;Z=? 

W  36      3*      81 


EXERCISE  101 

1.  Find  the  6th  term  of  1,  3,  9,  .... 

2.  Find  the  7th  term  of  6,  4,  f,  .... 

3.  Find  the  5th  term  of  -  2,  10,  -  50,  .... 

4.  Find  the  9th  term  of  3,  f,  |,  ••• 

5.  Find  the  10th  term  of  -•§,  +5,  -  10,  .... 

6.  Find  the  8th  term  of  %-,  ~,  f ,  .... 

32    lb    8 

7.  Indicate  the  11th  term  of  1,  (1  +  r),  (1  +  r)2, 


PROGRESSION'S  197 

8.  Indicate  the  15th  term  of  1}  i,  h  h    "*  a^so  tne  ^fcn 
term. 

Ttl        TYl       7YL 

9.  Indicate  the  13th  term  of  ra,  — ,  — ,  — ,  •••;  also  the 

o      9     21 

(n  +  l)th  term. 

10.  What  term  of  the  progression  3,  6,  12,  24,  ...  is  384  ? 

11.  What  term  of  the  progression  5,  10,  20,  •••  is  160? 

12.  What  term  of  the  progression  18,  6,  2,  •••  is  ^? 

13.  If  the  first  term  of  a  geometric  progression  is  5,  and 
the  6th  term  is  ^j  wna^  is  the  ratio  ? 

Find  the  ratio  of  the  geometric  progression  if: 

14.  The  first  term  is  -^  and  the  5th  term  is  -f. 

15.  The  first  term  is  f,  and  the  7th  term  24. 

176.  The  terms  of  a  geometric  progression  between  any  two 
other  terms  are  called  the  Geometric  Means  of  those  two  terms. 

Thus,  the  three  geometric  means  of  2  and  162  are  6,  18,  and  54,  since 
2,  6,  18,  54,  162,  form  a  geometric  progression. 

A  single  geometric  mean  of  two  numbers  is  particularly 
important.     It  is  called  The  Geometric  Mean  of  the  numbers. 

When  two  numbers  are  given,  any  specified  number  of  geo- 
metric means  may  be  inserted,  between  them. 

Example.     Insert  three  geometric  means  between  9  and  J^-. 

Solution  :  1.  There  results  a  geometric  progression  of  5  terms,  in 
which  a  =  9,  I  =  -x&^,  and  n  =  5.     Find  r. 

2.l  =  ar„-K     ...  |  =  9.r4,or^=15.     ,.,  =  ^?  =  ±|. 

3.   The  progression  is :  9,  6,  4,  f ,  ^-,  or  9,  —  6,  4,  —  |,  *$-. 
Check  :  There  is  a  G.  P.  with  three  terms  between  9  and  -1/. 


198  ALGEBRA 

EXERCISE  102 

1.  Insert  4  geometric  means  between  3  and  729. 

2.  Insert  5  geometric  means  between  2  and  128. 

3.  Insert  2  geometric  means  between  J  and  3. 

4.  Find  the  geometric  mean  of  8  and  32. 

5."  Find  the  geometric  mean  of  3  t  and • 

A.Z  t 

6.  Find  the  geometric  mean  between  2  x  and  8  x5. 

7.  Find  the  geometric  mean  between  —  and  —  • 

x  m 

8.  Find  the  geometric  mean  between  a  and  b. 

9.  Insert  3  geometric  means  between  3  and  12. 
10.   Insert  2  geometric  means  between  a  and  b. 

177.    The  Sum  of  the  First  n  Terms  of  a  Geometric  Progression. 

Given  the  first  term  a,  the  ratio  r,  and  the  number  of  terms 
n,  of  a  geometric  progression,  find  the  sum  of  the  terms  S. 
Solution:  1.    S  =  a  +  ar  +  ar2  +  •••  +  arn~2  +  arn_1.  (1) 

2.  Multiplying  both  members  of  (1)  by  r, 

rS  —  ar  +  ar2  +  ar3  +  •••  +  arn_1  -f  <B*«  (2) 

3.  Subtracting  equation  (2)  from  equation  (1), 

S—rS  =  a  —  arn,  or  #(1  —  r)  =  a  —  ar*.  (3) 

4.  .-.  S  =  ^=^!.  (4) 

5.  Since  ?  =  ar1"1,  then  rZ  =  arn.    Substituting  rl  for  arn  in  equation 

Example.     Find  the  sum  of  the  first  6  terms  of  2,  6,  18  •••. 
Solution:  1.    a  =  2,  r  =  3,  w  =  6.     Find  & 

fl  a  2-2-36  _  2  -  1458  =  -1456  =  72g 
1_3  _2  -2 


PROGRESSIONS  199 

EXERCISE  103 
Find  the  sum  of  the  first : 

1.  Eight  terms  of  the  progression  5,  10,  20,  ••.. 

2.  Six  terms  of  the  progression  24,  12,  6,  •  ••. 

3.  Seven  terms  of  the  progression  5,  —  15,  -+-  45,  ••<> 

4.  Seven  terms  of  the  progression  -Jg,  —  |,  i,  .... 

5.  Five  terms  of  the  progression  —2,  10,  —50,  •••. 

6.  Fifteen  terms  of  the  progression  3  m,  3  m3,  3  m5,  •••. 

7.  Ten  terms  of  the  progression  1,  m2,  m4,  m6,  .... 

8.  Find  the  sum  of  15  terms  of  1,  (1  4-  r),  (1  +  rf,  .... 

9.  Find  the  sum  of  the  first  10  powers  of  2. 

10.  Find  the  sum  of  the  first  10  powers  of  3. 

11.  Each  year  a  man  saves  half  as  much  again  as  he  saved 
the  preceding  year.  If  he  saved  $128  the  first  year,  to  what 
sum  will  his  savings  amount  at  the  end  of  seven  years  ? 

12.  Find  the  sum  of  the  terms  from  the  11th  to  the  15th 
inclusive  in  the  progression  y1^,  \,  \,  «... 

13.  A  father  agrees  to  give  his  son  5  ^  on  his  fifth  birthday, 
10^  on  his  sixth,  and  each  year  up  to  the  21st  inclusive  to 
double  the  gift  of  the  preceding  year.  How  much  will  he 
have  given  him  altogether  after  his  21st  birthday? 

178.  Infinite  Geometric  Progression.  By  an  infinite  geo- 
metric progression  is  meant  one  the  number  of  whose  terms 
increases  indefinitely.  If  the  ratio  is  greater  than  one,  the 
terms  become  larger  and  larger.  For  example,  the  progression 
3,  6,  12,  24,  •••.  If  Sn  represents  the  sum  of  the  first  n  terms 
of  a  progression,  then,  when  r  is  greater  than  1,  Sn  increases  in- 
definitely as  n  increases  indefinitely. 

Thus,  in  the  progression  3, 6, 12,  •••,  as  n  increases  indefinitely, 


200 


ALGEBRA 


Sn  increases  indefinitely.  Hence  the  sum  of  an  infinite  num- 
ber of  terms  of  the  progression  must  be  an  indefinitely  large 
number. 

When  the  numerical  value  of  the  ratio  is  less  than  1,  the 
progression  has  special  interest. 

Example  1.     Consider  the  progression  5,  -f,  f,  .... 

Solution  :    1.    The  ratio  r  is  |. 

2. 


When 
n  is  : 

l  =  at—1 

is: 

n      1-r 
is : 

4 

5Q)3  =  A 

10 

KW  =  iwhi 

1-i                       1-i 

3.  Clearly,  as  n  increases,  I  decreases  ;  also  the  terra  rl  of  Sn  decreases. 
If  11  increases  indefinitely,  I  will  become  approximately  zero,  the  term  rl 
will  become  approximately  zero,  and  Sn  will  become  approximately 

5     _5     35 
1-1      t       2* 
Consider  now  any  geometric  progression  in  which  r  is  less 
than  1  in  absolute  value  (§  21).     The  sum  of  the  first  n  terms 
1S :  o      a  —  arn 

1  —  r 

Now  as  n  increases  indefinitely,  rn  decreases  indefinitely, 
becoming  approximately  zero.  Hence  the  term  a  •  rn  becomes 
approximately  zero,     a,  and  1  —  r  remain  the  same. 

.•.  Sn  becomes  approximately     ~     or  • 

Hence,  the  sum  of  an  infinite  number  of  terms  of  a  geo- 
metric progression  in  which  r  is  numerically  less  than  1,  is 

given  by  the  formula  S  = 

1  —  r 


PROGRESSIONS  201 


Example.     Find  the  sum  to  infinity  of  the  progression 

8     16     ... 


4,  -A 


Solution:    1.    a  =  4;r==—  |. 

2.    Since  r  is  numerically  less  than  1 ,  S  =  — - — 

1  -  r 


1  +  1 


3 


EXERCISE  104 
Find  the  sums  to  infinity  of : 

6.    x, 


1.  6,2,1,....  fi    mx  ? 

2'  4' 


9      1      1      1     ... 

"  *    a.  * ■■'-•. 

3.   16,  4,  1,  ....  '     '  10'  100'  '"• 

4    5    I      «     ...  8-   1>~h+^,-- 

*•     u>   lTF>   lT0">         *  q_5_10  20 

W*  3?  ~9  >  2T  >         * 

5.    1,  .1,  .01,  .001,  ....  10.    t-^+A...... 

11.  Find  the  value  of  the  repeating  decimal  .8181  .... 

Solution:   1.    .8181  ...  =  fa  +  fa^  +  etc.  .... 

2.  This  is  a  G.  P.  in  which  a  =  fa  ;  r  =  j^-  The  value  of  the 
decimal  if  an  infinite  number  of  decimal  places  is  considered  is  given  by 
the  formula 

S=-^-  (§296). 
l  —  r 

.  8=     ^     =81  x100  =  81=9 
1~xk     10°      "      "     n 

Find  the  values  of  the  following  repeating  decimals : 

12.  .3333  ....  14.   .5333  ....  16.    .212121  .... 

13.  .7777-...  15.    .6444-...  17.   .151515-... 


XVII.    THE   BINOMIAL   THEOREM 

179.  The  Binomial  Theorem  is  a  formula  for  determining  by 
inspection  the  expansion  of  any  power  of  a  binomial. 

By  actual  multiplication: 

(a  +  xf  =  a2  +  2  ax  +  x2.  (1) 

(a  +  xy  =  a3  +  3a2x  +  3ax2  +  x3.  (2) 

(a  +  x)A  =  a4  +  4  A  +  6  a  V  +  4  aa3  +  x\  (3) 

Rule.  —  To  expand  any  power  of  a  binomial,  like  (a  +  jr)n : 

1.  The  exponent  of  a  in  the  first  term  is  n  and  decreases  by  1  in 
each  succeeding  term  until  it  becomes  1.  The  last  term  does  not 
contain  a. 

2.  The  first  term  does  not  contain  *.  The  exponent  of  x  in  the 
second  term  is  1  and  increases  by  1  in  each  succeeding  term  until  it 
is  n  in  the  last  term. 

3.  The  coefficient  of  the  first  term  is  1 ;  of  the  second  is  n. 

4.  If  the  coefficient  of  any  term  be  multiplied  by  the  exponent  of 
a  in  that  term,  and  the  product  be  divided  by  the  number  of  the 
term,  the  quotient  is  the  coefficient  of  the  next  term. 

Note  1.    The  number  of  terms  is  n  + 1. 

Note  2.  The  coefficients  of  terms  "equidistant  from  the  ends"  are  the 
same ;  for  example,  the  second  and  the  next  to  the  last. 

Example  1.     Expand  (a  +  x)5. 

Solution  :  1.  The  exponents  of  a  are  5,  4,  3,  2,  1.  The  exponents  of 
£,  starting  with  1  in  the  second  term,  are  1,  2,  3,  4,  and  5.  Writing  the 
terms  without  the  coefficients  gives : 

o5  +  a*x  +  as$  +  a2x3  +  a«*  +  xb. 

2.  The  coefficient  of  the  first  term  is  1,  and  of  the  second  term  is  5 
(Rule  3).     Multiplying  5,  the  coefficient  of  the  second  term,  by  4,  the 

202 


THE  BINOMIAL  THEOREM  203 

exponent  of  a  in  the  second  term,  and  dividing  by  2,  the  number  of  the 
term,  gives  10,  the  coefficient  of  the  third  term  ;  and  so  on. 
Filling  in  the  coefficients  in  this  manner  gives  : 

(a  +  x)5  =  a5  +  5  a4x  +  10  asx2  +  10  a?x8  +  5  as4  +  <e5. 
Example  2.     Expand  /2— Y. 
Solution  :  1.   In  this  example,  a  is  2  and  x  is  (  —  —  ]  • 

2.  .-.  ^2-|y  =  2»+6.2^-|Ul5.2*.(-|V+20.2».(-|)3 

+  15.22.(-f)V6.2.(-f)5+(-fy 

3.  =64-6-32.^+15.16.  ™?_20-8.  ™!+l5.4.  — 

3  9  27  81 

'  243      729 

,«         «,«      *m        ,  80     .      160     .  .  20  4      R  ,    m6 

4.  =  64  —  64  m  -\ m2 m3  +  —  m4 m6  H • 

3  27  27  81  729 

Note  1.  When  the  second  term  of  the  binomial  is  negative,  the  terms  of 
the  expansion  are  alternately  positive  and  negative. 

Note  2.  When  the  terms  of  the  binomial  are  complicated  monomials, 
place  each  in  parentheses,  and  afterwards  simplify  as  in  steps  3  and  4. 

EXERCISE  105 
Expand  the  following : 

1.  (x  +  y)4.  6.   (a2 -b^\  11.   (a-%y. 

2.  (m-n)\  7.    (2a  +  l)5.  12.    (i  +  x)\ 

3.  (c  +  1)4.  8.    (a-Sb)\  13.    (2m2-l)6. 

4.  (r-2)5.  9.    (1  +  z2)6.  14.    (a2-r-62c)4. 

5.  (m  +  n)\  10.    (l-x)\        '         15.    (3+ic8)5. 
Find  the  first  three  terms  of : 

16.  (a-3)15.  18.    (a-i)18.  20.    (aP  +  Stf)™ 

17.  (m2  +  2n)*°.  19.    (a2-^3)10.  21.   (m2-4w2)u. 


204  ALGEBRA 

23     (--«Y*  25.    (a"1  +  &~2)5.  27.    (V2-V3)6. 

28.   Write  the  first  4  terms  of  (a  +  x)M. 

180.  The  rth  or  General  Term  of  (a  +  x)n.  Following  the 
rales  of  §  297, 

(a  +  x)n  =  an  -f- fl  •  a71-1^  +  M71"1)  .  an" V 

J.  •  a 

.  n(n  —  l)(n  —  2)    _-  o  , 
+    v    lt2.3 +  "* 

Note  the  fourth  term.  The  exponent  of  a;  is  1  less  than  the 
number  of  the  term ;  the  exponent  of  a  is  n  minus  the  expo- 
nent of  x ;  the  last  factor  of  the  denominator  equals  the  expo- 
nent of  x ;  in  the  numerator  there  are  as  many  factors  as  there 
are  factors  in  the  denominator.     Hence, 

Rule.  —  In  the  rth  term  of  (a  +  x)n : 

1.  The  exponent  of  x  is  r  -  1. 

2.  The  exponent  of  a  is  n  —  the  exponent  of  x,  i.e.,  n  —  r  + 1. 

3.  The  denominator  of  the  coefficient  is  1 .2*3  •••  (r— 1),  the 
last  factor  being  the  same  as  the  exponent  of  x. 

4.  The  numerator  of  the  coefficient  is  n(n  —  1)  •••  etc.,  until  there 
are  as  many  factors  as  in  the  denominator. 

.-.  The  rth  term  is  n(n- 1)  ...  (n  -  r+ 2)  ,  fln_r+1 . ^ 
1-2...  (r-1) 

Example.     Find  the  8th  term  of  (3  c$  -  b)l\ 

2.   Solution:  1.    (3  at  -  b)n  ={(3  a-)  +  (-  6)}11. 

In  the  8th  term,  the  exponent  of  (—  b)  will  be  7  (Rule  1);  the  ex- 
ponent of  (3  a?)  will  be  11  —  7,  or  4  ;  the  last  factor  of  the  denominator 
will  be  7,  and  there  will  be  7  factors  in  the  numerator  starting  with 
11-10,  etc. 


u 


THE   BINOMIAL   THEOREM  205 

3 
3.    ...  The  8th  term  is  )1^1^I^1 .  (3  a*)*(-  &)7, 

or  330(81  a2)  (-  &»)  m  -  26730  a?W. 

Note.  If  the  second  term  of  the  binomial  is  negative,  it  should  be  in- 
closed, sign  and  all,  in  parentheses,  before  applying  the  rules.  Also,  if  either 
term  has  an  exponent  or  coefficient  other  than  1,  the  term  should  be  inclosed 
in  parentheses  before  applying  the  rules. 

EXERCISE  106 

Find  the : 

1.  4th  term  of  (a  +  xf.  8.  5th  term  of  (2  x2  -  3)10. 

2.  9th  term  of  (m-n)n.  9.  6th  term  of  (xm-yn)12. 

3.  5th  term  of  (a  +  2)°.  /_ 

10.  7th  term  of  --b 

4.  10th  term  of  (q  -  x)l\  \b 

5.  8th  term  of  (m*  -  *?.        n    ^  term  of  A?  _  ft* 

6.  6th  term  of  (a3 +  3  or5)10.  X*       XJ 

7.  7th  term  of  (c  -  I)15.  12.   8th  term  of  (or1  -  2  yl)» 

181.  The  Binomial  Formula  has  not  been  proved  in  this 
chapter;  it  has  been  written  down  from  observation  of  the 
results  in  certain  special  cases.  The  formula  has  been  ap- 
plied only  for  positive  integral  values  of  n. 

The  proof  of  the  formula  for  positive  integral  exponents  will 
be  found  in  §  219. 

In  more  advanced  courses  in  mathematics,  the  formula  is 
proved  to  be  correct  (with  certain  limitations)  not  only  for 
positive  integral  values  of  n  but  also  for  negative  and  frac- 
tional values. 

Historical  Note.    The  binomial  theorem  was  formulated  by  Newton. 


XVIII.    RATIO,    PROPORTION,    AND   VARIATION 

182.  The  Ratio  of  one  number  to  another  is  the  quotient  of 
the  first  divided  by  the  second. 

Thus,  the  ratio  of  a  to  b  is  -;  it  is  also  written  a  :  b.     The 

b 

numerator   is   called  the  Antecedent  and  the  denominator  is 

called  the  Consequent. 

All  ratios  are  fractions  and  are  subject  to  the  usual  rules  for 

operations  with  fractions. 

183.  The  ratio  of  two  concrete  quantities  may  be  found  if 
they  are  of  the  same  kind  and  are  measured  in  terms  of  the 
same  unit. 

Thus,  the  ratio  of  3  lb.  to  2  lb.  is  f  ;  and  the  ratio  of  350  lb.  to  2  tons  is 

EXERCISE  107 

Express  the  following  ratios  and  simplify  them : 

1.  3  to  9.      3.    5xto2x.         5.   f  to  ■&.      7.   25  to  375. 

2.  12  to  2.     4.   6  a2  to  15  a8.     6.    ■&  to  J.      8.    a2 - b2  to  a8- b\ 

9.   A  line  15  inches  long  is  divided  into  two  parts  which 
have  the  ratio  2 : 3.     Find  the  parts. 
Solution:  1.  Let  x  =  the  short  part. 
2.  Then  15  -  x  =  the  long  part. 

Complete  the  solution. 

10.   Divide  a  line  63  inches  long  into  two  parts  whose  ratio 

is  3 : 4. 

206 


RATIO,  PROPORTION,  AND  VARIATION  207 

11.  Divide  36  into  two  parts  such  that  the  ratio  of  the 
greater  diminished  by  4  to  the  less  increased  by  3  shall  be  3 :  2. 

12.  The  ratio  of  the  height  of  a  tree  to  the  length  of  its  shadow- 
on  the  ground  is  17 :  20.  Find  the  height  of  the  tree  if  the 
length  of  the  shadow  is  110  feet. 

13.  Divide  99  into  three  parts  which  are  as  2  :  3  :  4. 
Hint  :  Let  the  parts  be  2  x,  3  x,  and  4  x. 

14.  Divide  a  farm  consisting  of  720  acres  into  parts  which 
are  as  3  :  5. 

15.  Divide  $1000  into  3  parts  which  are  as  5  : 3  :  2. 

184.  A  Proportion  is  a  statement  that  two  ratios  are  equal. 
The  statement  that  the  ratio  of  a  to  b  is  equal  to  the  ratio  of 
c  to  d  is  written  either 

-  =  -,  or  a:  b  =  c:  a. 
b     d 

This  proportion  is  read  "  a  is  to  b  as  c  is  to  d.n 

Thus  3,  9,  5  and  15  form  a  proportion  since  f  =  j\. 

Historical  Note.  Leibnitz,  1646-1716,  was  instrumental  in  estab- 
lishing the  use  of  the  form  a  :  b  —  c  :  d. 

185.  The  first  and  fourth  terms  of  a  proportion  are  called 
the  Extremes,  and  the  second  and  third  the  Means. 

In  the  proportion  a :  b  =  c:  d,  a  and  d  are  the  extremes,  and 
5  and  c  are  the  means  ;  a  and  c  are  the  antecedents,  and  b  and 
d  are  the  consequents. 

EXERCISE  108 

Find  the  value  of  the  literal  number  in  the  first  six  of  the 
following  exercises  and  of  x  in  the  remaining  ones : 

1    ?_J1.  «7       c  „    2  —  x     5 


3     27  16     5  3         2 

y     10  24     z  '  i  +  t     2 


208 


a 

X 

7. 

—  ss 

;  — • 

b 

c 

a 

X 

8. 

: 

z^          ■« 

26 

3c 

ALGEBRA 

r2      r 

9.    —  =  -• 

sx      t 

11. 

a  — a;     a 
x         b 

10.    »=.«. 

np      nx 

12. 

a         n 

x—m     x 

186.  A  Mean  Proportional  between  two  numbers  a  and  b  is 

the  number  x  in  the  proportion  a  :  x  =  x :  b. 

2     a; 
A  mean  proportional  between  2  and  3  is  x  in  :  -  =  -  • 

X       o 

.*.  x2  =  6  j  a:  =  ±  V6. 

Thus,  there  are  two  mean  proportionals  between  any  numbers.     Gener- 
ally the  positive  one  is  used. 

187.  The  Third  Proportional  to  two  numbers  a  and  6  is  the 
number  x  in  the  proportion  a:b  =  b  :  x. 

2      3 

Thus,  the  third  proportional  to  2  and  3  is  x  in  :   -  =  - ; 

,\  2  X  as  9  and  x  =  4.5. 

188.  The  Fourth  Proportional  to  three  numbers  a,  b,  and  c  is 

the  number  x  in  the  proportion  a:b  =  c:  x. 

2      4 

Thus,  the  fourth  proportional  to  2,  3,  and  4  is  the  number  x  in  :  -  =  2 

o      #  J 
.-.  2  a;  =  12  and  x  =  Q. 

Note.    The  numbers  must  be  placed  in  the  proportion  in  the  order  in  which 
they  are  given,  as  in  the  illustrative  examples. 


EXERCISE  109 
Find  the  fourth  proportional  to: 

1.  2,  5,  and  4.  4.   35,  20,  and  141 

2.  5,  4,  and  2.  5.   6  a,  2  b,  and  c. 

3.  7,  3,  and  14.  6.   x,  y,  and  <cy. 

Find  the  mean  proportionals  between : 

7.  18  and  50.  9.   2  a  and  a. 

8.  2 \  and  f.  10.   12m2rc  and  3  win9. 


RATIO,   PROPORTION,   AND  VARIATION  209 


,,     a?-a-6       A  a2-\-a-\2  ,        ,       ,  af  +  xy  +  y2 

11.    : —  and  — I- 12.    Xs  —  y3  and  y      y  • 

a+4  a  +  2  *  a  —  y 

13-16.   Find  the  third  proportional  to  the  numbers  in  ex- 
amples 7,  S,  9,  and  10. 

17.  Find  the  third  proportional  to  a2  —  9  and  a  —  3. 

18.  Find  the  third  proportional  to  10  x  and  3  y. 

19.  Find  the  fourth  proportional  to : 

2  x2  —  2  y2     Xs  —  y8         ,ax  —  by  +  ay—bx 
a  +  b     '    P^d*'  aj«  +  ajy  +  y» 

20.  Find  the  mean  proportionals  between : 
ax-ay-bx  +  by  ^    tf-y* 


tf  +  xy+y2  (a  —  &)3 

PROPERTIES  OF  PROPORTIONS 

189.  In  a  proportion,  the  product  of  the  means  is  equal  to  the 
product  of  the  extremes. 

This  property  of  a  proportion  is  proved  as  follows : 

If  -  =-,  then  ad  =  be,  by  clearing  of  fractions. 
b      d 

Example.     Since  §  =  $ ,  2  .  9  should  equal  3  •  6.     Does  it  ? 

190.  If  the  product  of  two  numbers  is  equal  to  the  product  of 
two  other  numbers,  one  pair  may  be  made  the  means  and  the  other 
the  extremes  of  a  proportion. 

If  mn  =  xy,  then  —  =  &■  • 

x      n 

Prove  this  by  dividing  both  members  of  the  given  equation 

by  nx. 

Prove  that  the  following  proportions  also  are  true : 

(a)  £-5   (divide  by  ny).  (b)    *=*.  (c)  S^A. 

Example  1.     Since  3    8  =  6  •  4,  f  should  equal  f .    Does  it  ? 
Example  2.     Write  three  other  proportions  which  should  be  true  ac- 
cording to  the  property  given  in  this  paragraph. 


210  ALGEBRA 

191.  In  any  proportion,  the  terms  are  in  proportion  fr/ Alter- 
nation ;  that  is,  the  first  term  is  to  the  third  as  the  second  is  to  the 
fourth. 

xp  a     c  a     b 

If  -  =  -,  prove.  -  =  ;;• 

b     a  c     d 

Suggestion.     Use  §  189  and  then  divide  both  members  of  the  equation 

by  cd. 

Example.     Since  £  =  T^,  then  £  should  equal  T6j.    Does  it  ? 

192.  In  any  proportion,  the  terms  are  in  proportion  by  Inver- 
sion ;  that  is,  the  second  term  is  to  the  first  as  the  fourth  is  to  the 
third. 

If  -  =  -,  prove  -  =  - . 

b     d  a     c 

Suggestion.     Use  §  189,  and  then  divide  both  members  of  the  equation 

by  ac. 

Example.     Since  |  =  T\,  then  |  should  equal  \*.    Does  it  ? 

193.  In  any  proportion,  the  terms  are  in  proportion  by  Compo- 
sition ;  that  is,  the  sum  of  the  first  two  terms  is  to  the  second  as 
the  sum  of  the  last  two  terms  is  to  the  fourth. 

If  «  =  ^,  prove  a  +  b^c+d^ 

b     d*  b  d 

Suggestion.    Add  1  to  both  members  of  the  given  equation. 

Example.    Since  -  =  — ,  then  ■  +     should  equal  — — — .    Does  it  ? 

6      12  6  12 

194.  In  any  proportion,  the  terms  are  in  proportion  by  Divi- 
sion ;  that  is,  the  difference  of  the  first  two  terms  is  to  the  second, 
as  the  difference  of  the  last  two  is  to  the  fourth. 

Ha     c „n  a  —  bc  —  d 

-  =  -,  prove  ——  =  ——-. 

b     d  b  d 

Suggestion.     Subtract  1  from  both  members  of  the  equation. 

Example.     Since  —  =  — ,  then  10~2  should  equal  i^=^  •    Does  it  ? 
2       3  2  3 


RATIO,   PROPORTION,  AND  VARIATION  211 

195.  In  any  proportion,  the  terms  are  in  proportion  by  Compo- 
sition and  Division ;  that  is,  the  sum  of  the  first  two  terms  is  to 
their  difference  as  the  sum  of  the  last  two  terms  is  to  their 
difference. 

t-f  a     c   ™~™  a  +  5     c  +  d 

It  —  =  -,  prove  — ! —  =  — ! — • 

b     d  a—b     c—d 

Proof.    1.   Since  £  =  -,  then  2L±£  =  £+_!  (Composition) 

b     d  b  d 

2.  Since  frs^i  then  9-=A  =  °J=lA.  (Division) 

b     d  b  d  K  ' 

3.  Divide  the  members  of  the  equation  in  step  1  by  those  of  the  equa- 
tion in  step  2:  „  n      „      »>      ^ij      «      j 

fl  +  o  .  a  —  o  _c  +  a  .  c  —  a 

b  b  d  d 

a  +  b     c  +  d 


4.   Simplifying  step  3: 


b     c-d 


Example.     Since  —  =  — ,  then,  ™-±2  should  equal  15  +  3 .   Does  it  ? 
2       3'         '10-2  H       15-3 

196.  In  a  series  of  equal  ratios,  the  sum  of  the  antecedents  is  to 
the  sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 

If  2^*1,  etc.,  prove    a  ±  c  ±  e  ±  etc'  ,1 
b     d     f        '  *  6  +  d+/+etc.      b 

Proof.    1.  Let  v  =  the  common  value  of  the  equal  ratios  - ,  -,  -,  etc. 

b    d   f 

2.  Then  since  -  =  v,  a  —  bv. 

b 

-  =  v,  c  =  dv. 
d 

j  =  v,e=fv. 

3.  Then         (a  +  c  +  e)  =  bv  +  dv  +fv  =  v(b  +  d+f), 

*"**■»       b  +  d+f  b  +  d+f     b        d        f 

Example.    Since  1  =  -  =  —  ,     1+3  +  5    should  equal  -.    Does  it ? 
2     6      10     2+6  +  10  2 

Historical  Note.     All  of  these  properties  of  a  proportion  were  known 
to  Euclid,  300  b.c 


212  ALGEBRA 

197.  There  are  several  other  properties  of  a  proportion 
which  follow  directly  from  properties  of  an  equation  or  of  a 
fraction. 

(a)  If  -=-    then  —  =  —  •  Raise  both  members  to  the  third 

b      cC  63      d3  power. 

(K\  If  -  =  -     then  va  =  ^°  Extract  the   cube   root   of   both 

b~  d"            f/l      jfa  members. 

nc  Multiply  numerator  and  denomi- 

(c)  If  -  =  -,  then  — -  =  —  •  nator  of  the  first  ratio  by  m,  and 


b     d  mb      nd 

ma     mc 


(d)  If  -  =  -,  then 

b      d  nb       nd  equation  by  - 

lb 


of  the  second  by  n. 

Multiply   both   members   of   the 
m 


198.  In  the  preceding  paragraphs,  some  of  the  simple  prop- 
erties of  a  proportion  have  been  given.  There  are  many  others 
which  may  be  derived  by  means  of  these  simple  properties. 

-™  t*  a      c  2  a  4-  Sb     2a  —  Sb 

Example.     If  -  =  -,  prove  -^ — -=— -. 

b     d    F  2c  +  3d      2c -3d 

Proof.     1.   Since  2  =  -,  then  ?*  =?-S.  (§  197,  d) 

b      d  Sb      3d  vs  J 

2.  Then  |«±M  =  |£±|ij.  (By  §  m) 

2a  —  6b      2  c  —  6d 

3.  Then  2a  +  8  b  =  2  a  ~  3  h>  (By  §  191) 

2c+3d     2c-3d 


EXERCISE  110 

1.  Write  by  inversion : 

M  3     15  /7A  2     ra  ,  v   a     a? 

(a)  i=2-o-      (6)  rr      (c)  i=v 

2.  Write  these  same  three  proportions  by  alternation. 

3.  Write  these  same  three  proportions  by  composition. 

4.  Write  these  same  three  proportions  by  division. 

5.  Write  the  proportion  (c)  in  Example  1 : 

(a)  by  inversion  and  the  result  by  composition ; 


RATIO,  PROPORTION,   AND  VARIATION  213 

(6)  by  alternation  and  the  result  by  division ; 

(c)  by  composition  and  the  result  by  alternation ; 

(d)  by  division  and  the  result  by  inversion. 

6.  If  —  =  - ,  prove  that  — :L—  =  -  • 

n      y  x+y      y 

7.  If  -  =  -,  prove  that  —*—-=    —    • 

s     b  r  a 

T /?  a      c  ,-1    ,  a  —  b      b 

8.  If  -  =  - ,  prove  that  =  -  • 

b     d  c  —  d     d 

rt    T  o  x      w  ,  i    .  x2  +  u2     w2  -\-t2 

9.  If  -  =  — ,  prove  that  — ^- —  =  — 4—* 

u      t'  F  u2  t2 

tn    T£  a      c    „„„TTni.^n4.2a  —  3b     2c  — 3d 

10.   it  -  =  — ,  prove  that  = -• 

b     d}  p  b  d 


EXERCISE  111 
Proportion  in  Geometry 

1.  In  a  triangle  in  which  BE  is  parallel  to  BC,  m:r  =  n:s. 

To  test  this  truth  :  (a)  measure  wi,  n,  r,  and 
8 ;  (b)  find  the  value  of  the  ratio  m  :  r  and  of 
n  :  s  ;  (c)  compare  these  two  ratios. 

This  truth  may  be  tested  in  any  triangle.  It 
may  be  expressed  thus :  the  upper  segment  on 
one  side  is  to  the  lower  segment  on  that  side  as 
the  upper  segment  on  the  other  is  to  the  lower 
segment  on  the  other. 

2.  Write  the  proportion  —  =  -  by   alternation.     Express 
the  resulting  proportion  in  words  as  in  Example  1. 

3.  Write  the  proportion  of  Example  1  by  composition  and 
express  it  in  words. 

4.  Write  the  proportion  of  Example  1  by  inversion  and 
express  it  in  words. 

5.  If  AD  =  7,  DB  =  4,  and  AE  =  8,  find  EC. 


214 


ALGEBRA 


6.  If  AB  =  12,  AD  =  5,  and  AC=  14,  find  AE. 
Hint.     Let  AE  =  x,  and  CE  =  14  -  a. 

7.  If  ^4Z>  as  ZXB,  how  does  AE  compare  with  EC? 

8.  If  AD  =  20,  1)5  =  8,  and  AC=  30,  find  JLE  and  EC. 

9.  If  two  perpendicular  lines  BO  and 
Di?  are  drawn  from  one  side  of  an  angle 
to  the  other,  then  BC:AC=DE:  AE. 

Test  this  statement  by  measuring  the  lines 
in  the  figure  and  finding  the  value  of  the  ratios. 

10.  Draw  any  other  perpendicular,  as  XY.  Find  the  ratio 
of  XY  to  ^l^and  compare  the  ratio  with  those  found  in 
Example  9.     What  do  you   conclude 

about  all  ratios  obtained  by  dividing 
the  length  of  the  perpendicular  by  the 
distance  from  A  to  the  foot  of  the  per- 
pendicular (like  AY)? 

11.  Using  the   fact   stated  in  Ex-  ^i5-_- 

ample  9,  tell  how  to  find  the  height  • „ > 

of    the    tree    in    the    figure,    if    the 

height    of   the   rod   and  the  lengths  on  the  shadows  of  the 
tree  and  the  rod  are  as  indicated. 

12.  Suppose  that  EF  and  AC  are 
perpendicular  to  OC  in  the  adjoining 
figure.  Suppose  that  EF=  10  feet, 
OF  =12  feet,  0(7=150  feet,  and 
BO  as  20  feet.     Determine  AB. 

13.  Suppose  that  CD  and  AB  are 
perpendicular  to  AE  in  the  adjoining 
figure;  that  AX  =  5  feet,  YB  =  8 
feet,  AE=  750  feet,  CE  =  25  feet, 
and  CD  =  30  feet.     Eind  XY. 


RATIO,  PROPORTION,  AND  VARIATION  215 


VARIATION 

199.  Some  quantities  change  or  vary  and  are  called  Variable 
Quantities. 

Thus,  the  distance  between  a  moving  train  and  its  destination  varies,  — 
that  is,  it  decreases  j  the  age  of  an  individual  varies  from  moment  to 
moment,  —  that  is,  it  increases. 

200.  A  quantity  which  is  fixed  in  any  given  problem  is  called 
a  Constant. 

Thus,  if  a  workman  receives  a  fixed  sum  per  day,  the  total  wages  due 
him  changes  from  day  to  day  if  he  works  and  remains  unpaid.  His  daily 
wage  is  a  constant ;  his  total  wages  is  a  variable. 

201.  A  change  or  Variation  in  one  quantity  usually  produces 
a  variation  in  one  or  more  other  quantities.  Such  variables 
are  called  Related  Variables.  For  each  value  of  one  variable 
there  is  a  corresponding  value  of  the  other  variable,  or  variables. 

Thus,  if  the  side  of  a  square  is  increased,  the  perimeter  and  the  area  of 
the  square  are  also  increased. 

202.  Variation  is  the  study  of  some  of  the  laws  connecting 
related  variables.  Instead  of  the  quantities  themselves,  their 
measures  in  terms  of  certain  units  of  measure  are  used. 

Thus,  distance  is  expressed  as  a  number  of  miles,  rods,  or  other  units 
of  length  ;  weight  is  expressed  as  a  number  of  units  of  weight ;  area  is 
expressed  as  a  number  of  units  of  area. 

203.  One  quantity  varies  directly  as  another  when  the  ratio 
of  any  value  of  the  one  to  the  corresponding  value  of  the  other 
is  constant. 

Thus,  the  ratio  of  the  perimeter  of  a  square  to  the  side  of  the  square  is 
always  4,  because  the  perimeter  is  4  times  the  length  of  the  side ;  there- 
fore the  perimeter  varies  directly  as  the  side  of  the  square. 

204.  The  symbol,  oc,  is  read  "  varies  as  " ;  thus,  a  oc  b  is  read 
"  a  varies  as  b." 


216  ALGEBRA 


x 

If  xazy,  then  -  =  m,  where  m  is  a  constant,  expresses  the 

relation  between  any  two  corresponding  values  of  x  and  y. 
(See  §  203.) 

x 
Since  -  =  m,  then  x  =  my. 

Either  equation  may  be  used  to  express  direct  variation. 


205.  One  quantity  is  said  to  vary  inversely  as  another  when 
the  product  of  any  value  of  the  one  and  the  corresponding 
value  of  the  other  is  constant. 

Thus,  the  time  and  rate  of  a  train  going  a  distance  d  are  connected  by 
the  equation  rt  =  d.  If  the  distance  remains  fixed,  then  the  time  varies 
inversely  as  the  rate  ;  for  example,  if  the  rate  is  doubled,  the  time  is  halved. 

If  x  varies  inversely  as  y,  then  xy  =  m,  where  mis  a  con- 
stant, expresses  the  relation  between  them. 

If  xy  =  m,  then  also  x  =  — .     Either  equation  may  be  used 

to  express  inverse  variation. 

206.  One  quantity  is  said  to  vary  jointly  as  two  others 
when  it  varies  directly  as  their  product.     If  x  varies  jointly 

as  y  and  z,  then  —  =  m,  where  m  is  a  constant,  expresses  the 

relation  between  the  variables. 

Thus,  the  wages  of  a  workman  varies  jointly  as  the  amount  he  receives 
per  day  and  the  number  of  days  he  works ;  for,  letting  W  equal  his  total 
wages,  w  his  daily  pay,  and  n  the  number  of  days  he  works,  then  W—nw. 
Here  m  =  1. 

Again,  the  formula  for  the  area  of  a  triangle  is 
A  =  I  ab. 

This  shows  that  the  area  of  a  triangle  varies  jointly  as  the  base  and 
altitude.     (Here  m  =  J.) 


RATIO,  PROPORTION,    AND  VARIATION  217 

207.  One  quantity  may  vary  directly  as  a  second  and  inversely 
as  a  third.     Let  x  vary  directly  as  y  and  inversely  as  z ;  then 

mil 

#  =  — 
z 

expresses  the  relation  between  the  variables.  Notice  that  this 
combines  the  equation  for  direct  variation  of  y  and  inverse 
variation  of  z. 

208.  Variation  of  more  complicated  related  variables  needs 
to  be  expressed  sometimes. 

Example  1.     x  oc  y2  may  be  written  x  =  my2. 

Example  2.     a?ccy2  may  be  written  xs  =  my2. 

Example  3.  The  volume  of  a  circular  cylinder  varies  jointly 
as  the  altitude  and  as  the  square  of  the  radius.  This  may  be 
expressed  :  v  oc  ar2,  or  v  =  har2 

Example  4.     a  varies  directly  as  q,  and  inversely  as  d\ 

This  may  be  expressed :    a  =  -f  • 

EXERCISE  112 

Express  the  following  relations  both  by  means  of  the  symbol  oc 
and  by  an  equation  : 

1.  The  area  of  a  rectangle  varies  jointly  as  the  base  and 
altitude. 

2.  The  area  of  a  circle  varies  as  the  square  of  the  diameter. 

3.  The  volume  of  a  rectangular  prism  varies  jointly  as  the 
length,  width,  and  height. 

4.  The  distance  a  body  falls  from  a  position  of  rest  varies 
as  the  square  of  the  number  of  seconds  in  which  it  falls. 

5.  The  interest  varies  jointly  as  the  principal,  the  rate,  and 
the  time. 


218  ALGEBRA 

Express  the  following  relations  by  means  of  equations: 

6.  The  rate  of  a  train  varies  inversely  as  the  time,  if  the 
distance  is  constant. 

7.  The  rate  of  gain  varies  inversely  as  the  capital  invested, 
if  the  total  gain  is  constant. 

8.  The  weight  of  an  object  above  the  surface  of  the  earth 
varies  inversely  as  the  square  of  the  distance  from  the  center 
of  the  earth. 

9.  The  per  capita  cost  of  instruction  for  pupils  in  a  school 
room  varies  directly  as  the  salary  of  the  teacher  and  inversely 
as  the  number  of  the  pupils. 

10.  The  volume  of  a  circular  cone  varies  jointly  as  the  alti- 
tude and  the  square  of  the  radius. 

11.  If  z  varies  jointly  as  x  and  y,  and  equals  §  when  y  =  -| 
and  x  =  f ,  find  z  when  y  =  £  and  x  —  \. 

Solution.     1.   According  to  the  conditions  z  =  mxy. 

2.  .-.  -  =  m  •  -  •  * ,  or  m  =  -,  since  z  —  \  when  x  =  f  and  y  =  f. 

6^5  3 

3.  .'.0  =  |  xy,  substituting  §  for  m. 

4.  .-.  z  =  |  •  |  •  f  =  ^,  when  x  =  f  and  y  =  \. 

Note.    In  such  problems,  first  find  the  constant,  as  in  step  2. 

12.  If  y  oc  x  and  is  equal  to  40  when  <c  =  5,  what  is  its  value 
when  x  =  9  ? 

13.  If  yocx*  and  is  equal  to  40  when  sc  =  4,  what  is  the 
equation  for  y  in  terms  of  #? 

14.  If  x  varies  inversely  as  y  and  is  equal  to  -|  when  y  =  f , 
what  is  the  value  of  y  when  x  is  |  ? 

15.  If  (5  a;  +  8)  oc  (6  y  —  1)  and  #  =  6  when  ?/  =  —  3,  what  is 
the  value  of  a;  when  y  =  7? 

16.  The  distance  fallen  by  a  body,  from  a  position  of  rest, 
varies  as  the  square  of  the  number  of  seconds  in  which  the 


RATIO,  PROPORTION,   AND  VARIATION  219 

body  falls.     If  it  falls  256  feet  in  4  seconds,  how  far  will  it 
fall  in  6  seconds  ? 

17.  The  interest  on  a  sum  of  money  varies  jointly  as  the 
rate  of  interest  and  the  principal.  If  the  interest  is  $375 
when  the  rate  is  5  %  and  the  principal  is  $  3000,  what  is  the 
interest  when  the  rate  is  6  %  and  the  principal  is  $  2500  ? 

18.  The  principal  varies  directly  as  the  interest  and  inversely 
as  the  rate.  If  the  principal,  $4000,  produces  $250  interest 
at  4%,  what  principal  must  be  invested  for  the  same  time  to 
yield  $500  at  5%  ? 

19.  The  number  of  tiles  required  to  cover  a  given  area 
varies  inversely  as  the  length  and  width  of  the  tile.  If  it  takes 
270  tiles  2  inches  by  5  inches  in  size  to  cover  a  certain  area, 
how  many  tile  3  inches  by  6  inches  will  be  required  for  the 
same  area  ? 

20.  The  number  of  posts  required  for  a  fence  varies  inversely 
as  the  distance  between  them.  If  it  takes  80  posts  when 
they  are  placed  12  feet  apart,  how  many  will  be  required 
when  they  are  placed  15  feet  apart  ? 


XIX.    GRAPHICAL    REPRESENTATION    OF    COMPLEX 

NUMBERS 

209.  Representation  of  Real  Numbers.  Mark  any  point,  0, 
on  a  straight  line  X'X  by  the  number  0,  and  any  other  point, 

^ ^  U,  to  the  right  of  0  by  the  num.- 

•'  \  ber  -f-1.     Let  OU  be  considered 

/  \  the  unit  length. 

x/    ,  4A,      f      °     y     ,  ^A.     x        Any  real  positive  number,  +  a, 
-a-      -i      o    +i    4  2+a+ 3  »a   represent;e(j   Dy  i^q    p0int  A, 

a  units  to  the  right  of  0,  and  any  real  negative  number,  —  «, 
by  the  point  A',  a  units  to  the  left  of  0. 

210.  The  representation  of  —  a  (the  point  A')  may  be 
located  by  turning  the  representation  of  +  a  (the  point  A) 
about  the  point  0  as  center,  through  two  right  angles  in  the 
direction  opposite  to  the  motion  of  the  hands  of  a  clock. 

—  a  —  {-\-a)  x(  —  1),  hence  the  multiplier  —1  may  be  re- 
garded an  operator  which  turns  the  representation  of  +  a 
through  two  right  angles  in  the  counter-clockwise  direction, 
about  point  0  as  center. 

211.  Graphical  Representation  of  Pure  Imaginaries.  By 
definition  (§  82),  tx*  =  — 1.  Since  multiplication  of  +  a  by 
—  1  turns  the  representation  of 

4-  a  through  two  right  angles  in 
the  counter-clockwise  direction, 
i  may  be  regarded  an  operator 
which  turns  the  representation 
of  +a  through  one  right  angle 
in  the  counter-clockwise  direc- 
tion. This  suggests  represent- 
ing +  ai  by  the  point  B,  a  units 
above  0  on  YY'. 

220 


/ 
/ 
I 

Y 

> 
B 

s 

\ 
\ 

I* 

O 

iA 

-a 

0 

+  a 

-ai« 

b' 

GRAPHICAL   REPRESENTATION    OF   NUMBERS      221 


In  general,  pure  imaginaries  are  represented  by  points  on 
the  line  YY'.  ai  is  represented  by  the  point  a  units  above  0, 
and  —  ai  by  the  point  a  units  below  0.  YY'  is  called  the 
axis  of  pure  imaginaries. 


212.  Graphical  Repre- 
sentation of  Complex  Num- 
bers.  To  represent  a + bi : 

Let  A  represent  the 
number  a,  and  B  the 
number  bi.  Draw  AC 
equal  and  parallel  to  OB. 
Then  it  is  agreed  to  con- 
sider point  C  the  repre- 
sentation of  a-\-bi.  Thus 
D  represents   —  4  —  3  i. 


■5   -4    -3    -2    -I 


0« 


-2i 
— 3i  t 

-4i 

I 

Y 


I 


I     +2     +3    +4    +5 


7.  2.5  +  ?'. 

8.  4-2.5*. 


EXERCISE   113 
Represent  on  a  diagram  the  numbers : 

1.  2  +  *.         3.    -4  +  2t.         5.    -3  +  2*. 

2.  2-3  i,      4.    -3-2*.         6.    3-5L 
9.    Represent  a  +  &a  and  —  (a  -j-  fri). 

213.  Graphical  Represen- 
tation of  Addition  of  Complex 
Numbers. 

(a)  To  represent  graphi- 
cally the  sum  of  a  -+-  6*  and 
c  +  cfa*. 

Let  M  represent  a  -{-bi 
and  AT,  c  +  di. 

Construct  OJlf  and  ON, 
and  then  construct  the  par- 
allelogram OMPN,  thus 
locating  point  P. 


Y 

di 

— r 

bi 
0 

t  '        1 

c               a             X 

y' 

222 


ALGEBRA 


Point  P  represents  (a  4-  bi)  4  (c  4  di). 

Point  P  may  be  determined  readily  without  constructing  the 
complete  parallelogram,  by  drawing  from  M  the  line  MP  equal 
and  parallel  to  ON,  thus  adding  (c  4-  di)  to  (a  4  bi). 

Note.  The  correctness  of  the  construction  is  readily  proved  by  plane 
geometry.     The  proof  is  omitted  from  the  text. 

EXERCISE   114 

Represent  graphically  the  sum  of : 


3  +  i  and  2  +  5  i. 
2  -  3  i  and  1  4  4  L 
2  +  4  i  and  5  —  i. 
-  6  +  2  i  and  -  4  -  7 


5.    3  and  4 


Y 

bi 

f' 

o 

1 

j-a 

0                  a 
Y' 

6.  —  5  t  and  6  -f-  2  & 

7.  —  3  +  4  i  and  +  5. 

8.  4  5  —  3  i  and  —  4  & 

214.  Graphical  Represen- 
tation of  Subtraction  of 
Complex  Numbers.  In  the 
adjoining  diagram,  if  M 
represents  the  number 
a  +  bi,  then  M'  represents 
—  (a  4  bf),  for  M'  repre- 
sents the  number  —  a  —  bi. 
To  subtract  (a  4  bi)  from  (c  4  di),  one  may  add  —  (a  4  bi)  to 
(c  4-  di).     To  represent  this  difference  graphically : 

1.  Locate  M  representing  a  +  bi 
and  M '  representing  —  a  —  6i. 

2.  Locate  JV  representing 
c  4  d£. 

3.  Draw  from  N  a  line 
equal  and  parallel  to  OJ/7, 
thus  locating  R.  This  con- 
struction represents  adding 
-(a+W)  to  (c+di).  (§213.) 

4.  i£  represents  c4-^0 
-(a4&0- 


.  GRAPHICAL   REPRESENTATION   OF   NUMBERS     223 

EXERCISE   115 

1-8.  Represent  graphically  the  result  of  subtracting  the 
second  number  from  the  first  number  in  Examples  1-8  of 
Exercise  114. 

215.  It  is  possible  to  represent  graphically  other  operations 
with  complex  numbers,  but  such  topics  are  beyond  the  scope 
of  this  text. 


XX.   EQUATIONS   IN   THE   QUADRATIC   FORM 

216.    An  equation  is  in  the  quadratic  form : 

1.  if  it  has  three  terms ; 

2.  if  two  of  the  terms  contain  the  unknown  number ; 

3.  if  the  exponent  of  the  unknown  number  in  one  of  these 
two  terms  is  twice  its  exponent  in  the  other. 

Note.     The  unknown  number  may  be  an  algebraic  expression. 
Example  1.     Solve  the  equation  16  x*  —  22  af  *  —  3  =  0. 

_3  _3 

Solution  :  1.   Let  y  =  x  ¥  and  therefore  y2  =  x  2. 

2.  Hence  the  equation  becomes  16  y2  —  22  y  —  3  =  0. 

3.  .-.  (8y  +  l)(2y-3)=0. 

4.  .VJf=— 1,  orff  =  §; 

_3  _3 

that  is  x  ?  =  —  £,  or  x  ?  =  § . 

5.  From  x~*  =  ->,  x~*  =  V(^fj. 


If  the  principal  cube  root  (§  117)  of  —  4  is  taken,  x  = =  —  =  16, 

There  are  also  two  imaginary  roots,  obtained  by  taking  the  fourth  powers 
of  the  two  imaginary  cube  roots  of  ( —  |) .     (§  95.) 

6.   From  x~*  =  f ,  af  *  =  \/(|). 

x  again  has  one  real  value  and  two  imaginary  values. 

Altogether  there  are  6  roots  for  the  equation ;  the  principal  roots  are 

16  and  (v^)4. 

224 


EQUATIONS  IN   THE   QUADRATIC   FORM  225 

EXERCISE   116 
Solve  the  equations : 

1.  a4  -  29  a2  =  -  100.  6.  2s-8-3os-4  +  48  =  0. 

2.  27s6  +  46a8-16  =  0.  7.  67i-2  =  llV/L 

3.  16  a8 -33 a4- 243  =  0.  8.  a;*  +  33a;$  =  -32. 

4.  161  a5 +  5 +  32  a10  =  0.  9.  ^-10^  +  9  =  0. 

5.  3  x-'2  +  14  x-1  =  5.  10.  2a  +  3Vz  =  27. 

217.  An  equation  may  sometimes  be  solved  with  reference 
to  an  expression,  by  regarding  the  expression  as  the  unknown 
number. 

Example  1.    Solve  the  equation 


x2  -  6  x  +  5  V x2  -  6  x  +  20  =  46. 


Solution  :  1.  Let  y  =  Vx2^6~aT+  20, 

and  therefore  y2  =  x2  —  6  x  +  20. 

2.  The  equation  becomes 

y2  +  5  ?/  =  66,  or  */2  +  5  y  -  66  =  0. 

3.  .-.  (?/  +  ll)(j/-6)  =  0,  or  */  =  -ll  and  y  =  6. 


4.  When  ?/  =  6,  Vx2  -  6  x  +  20  =  6. 

.-.  x2  -  6  a;  +  20  =  36,  x2  -  6  x  -  16  =  0. 
.-.  (x-8)(x  +  2)  =  0,  or  x  =  8  and  x=-  2. 

5.  When  y  =-  11,  Vx2  -  6x  +  2~0  =-  11. 

.  •.  x2  -  6  x  +  20  =  121,  or  x2  -  6  x  -  101  =  0. 

^=6±V36+404  =  6iV410=6±2ViT0:=3        ^ 
2  2  2 


Note .     If  ?/,   or  Vx2  —  6  x  +  20,  is  restricted  to   the  principal  root, 
these  last  two  values  are  not  admissible. 


226  ALGEBRA 

EXERCISE  117 
Solve  the  following  equations  : 
1.    (2x2-3xy-$(2xi-3x)=9. 
(Hint  :  Let  y  -  2  x2  -  3  as.) 


2.   5  x  + 12  +  5  V5  x  + 12  =  -  4. 
(Lety  =V5z  +  12.) 

'      2a;    +a^-3  4  * 


d2  +  2      2  d  -  5     35 


5. 


2d-5      d2  + 


6.  arJ  +  7Va^-4a;  +  ll=4a;-23. 

7.  V»i2-3m-3  =  m2-3m-23. 
?-5t  +  l      t2-2t  +  2         8 


8. 


?-2t  +  2     t2-5t  +  l 


9.   2r2  +  4r  +  Vr2  +  2r-3  =  9. 


10.  c2  +  l-fVc2-8c  +  37  =  8(c  +  12). 

11.  25(a;  +  l)-1-15(a;  +  l)^  =  -2. 


-  #?->P 


2/2  +  3     2 


XXI.   THE   BINOMIAL   THEOREM    (Continued) 

218.  The  Binomial  Theorem  was  formulated  in  general 
form  (for  positive  integral  exponents),  in  §  179,  after  special 
cases  of  the  general  theorem  were  exhibited.  The  theorem 
was  not  proved;  it  was  arrived  at  by  the  process  of  pure 
induction. 

219.  Proof  of  the  Binomial  Theorem  for  Positive  Integral 
Powers.     Assume,  as  in  paragraph  179,  that 

(a+x)n=an+nan~lx  ±n(n-l\»-2x*  +  n(n-l)(n~2Kn-*3<*+ ...      (1) 

1  •  2  1  •  a  •  o 

Multiply  both  members  of  (1)  by  a+x.     Then 

(a  +  X)n+l^an+I  +  mna.+  ^(w-l)an-lx2+^-1Xn-2)aW_2a;8+       _ 
I    ■  -  1  •  It  •  o 

+  anx  +  jan~W  +  wfo-l)  an-2&  +  .... 

1  1  •  '_ 

.-.  {a  +  x)n+l  =  an+l  +  {n+\)anx+n\1?^+\\in-1x2 

1.2       L    3  J 

=  an+l  +  (n+  l)anx  +  n  ,  «±1  ^n-la-2  +  n(ft-l)  >  rc±Ian-2;K3  +  _ 

2  1-2  o 

=  a«+i+  („+!)„"*+  (w  +  1)-wa"-^+  (w  +  1)-n-(w-iya>^+  .-. 
1*2  1  •  2  •  o 

It  will  be  observed  that  the  expansion  on  the  right  is  in 
accordance  with  the  rules  of  §  179.     This  proves  that  if  the . 
rules  of  §  179  are  assumed  for  any  particular  positive  integer, 
n,  they  hold,  true,  also,  for  the  next  greater  integer,  n  -f- 1. 

227 


228  ALGEBRA 

But  the  rules  are  known  to  be  satisfactory  in  the  case  of 
(a  +  xy-,  hence  they  hold  for  (a+#)5.  Since  they  hold  for 
(a  +  x)b,  then  they  hold  also  for  (a  +  x)6 ;  and  so  on. 

Therefore  the  binomial  theorem  is  true  for  any  positive 
integer. 

Note.     The  above  method  of  proof  is  known  as  mathematical  induction. 

220.  Fractional  and  Negative  Exponents.  In  the  expan- 
sion of  §  219,  if  n  is  a  positive  integer,  there  is  ultimately 
a  term    (the   (n  +  2)nd),  for  and  after  which  the  coefficient 

n(n  —  l)(n-2)  •  ••   • 

™ & - —  is  zero. 

1.2-3... 

When  n  is  a  negative  integer  or  a  fraction,  there  is  no  term 
for  which  the  coefficient  is  zero.  Hence  the  terms  continue 
indefinitely.  The  resulting  expansion  has  an  infinite  number 
of  terms. 

In  this  case  also,  the  expansion  on  the  right  in  §  219  has 
a  sum,  and  this  sum  is  (a  +  x)n  for  any  rational  (§  112)  value 
of  71,  provided  the  absolute  value  of  a  is  greater  than  the 
absolute  value  of  x.  This  theorem  is  proved  in  a  more  ad- 
vanced course  in  mathematics. 

Assuming  the  theorem,  the  following  examples  may  be 
solved : 

Example  1.     Expand  (a  +  x)~*  to  four  terms. 

Solution  :  1.   Substitute  §  for  n  in  the  formula. 

vy  3  1-2  1.2-3 

=  0I+  2a~W  f(-i)a-fX2+  KHK-f) .  a-jX3+  ... 

o  2i  1  •  52  •  o 

=  J -ff  a~%x-\  a~^2+/T«~3^8+  •••• 

Example  2.     Find  the  7th  term  of  (a  —  3  &\)~t. 

Solution:  1.  The  7th  term  may  be  found  by  applying  the  rule  in 
§  180. 


THE    BINOMIAL   THEOREM 


229 


2.    Substitute  (—  3  x  2)  for  at,  and  (-  |)  for  n. 


The  exponent  of  (—  3x  z)  is  7  -  1  or  6. 
The  exponent  of  a  is  —  ^  —  0  or  —  Xf . 
The  denominator  of  the  coefficient  is  1  •  2  •  3  •  4  •  5  •  6. 
The  numerator  of  the  coefficient  is  (—  j).{— J  —  1)...  until  there  are 
six  factors. 

Hence  the  seventh  term  is : 

(-^(-#)(-l)(-^)(-¥)(-V)  (a-V)(_  3  afty 
1.2-  3-4-  5-  6  v         JK  J 

38  9 


EXERCISE   118 

Find  the  first  four  terms  of : 

1.  (a  +  x)K 

2.  (1+z)-8. 

3.  (l~x)~\ 


4.    V( 


b. 


Find  the 
9.    6th  term  of  (a  +  a;)i 

10.  5th  term  of  (a  —  &)"&. 

11.  7th  term  of  (1  -f  x)~7. 

12.  8th  term"  of  (1  —  xft. 


6.  (a* +  2  6)*. 

7.  (a3  -  4  a;*)*. 

8.  _!_. 

13.  9th  term  of  (a-a)'3. 

14.  11th  term  of  V(m  +  w)5. 

15.  7th  term  of  (cr2-2  &*)- 

16.  8th  term  of — . 


221.  Extraction  of  Roots.  The  Binomial  Theorem  may  some- 
times be  used  to  find  the  approximate  root  of  a  number  which 
is  not  a  perfect  power  of  the  same  degree  as  the  index  of  the 
root. 

Example.     Find  V'25  approximately  to  five  decimal  places. 

Solution  :  1.    The  nearest  perfect  cube  to  25  is  27. 


230  ALGEBRA 


2.    .-.  v/25  =  ^/27-2=[(33)  +  (-2)]^ 

=  (3B)^+i(33)-|(_2)-|(33)-f(_2)2+A(33)-t(_2)3... 


=3 


3  •  32      9  •  35     81  •  38 
=  3  -  .07407  -  .00183  -  .00008  •••  =  2.92402. 

Rule.  —  Separate  the  given  number  into  two  parts,  the  first  of 
which  is  the  nearest  perfect  power  of  the  same  degree  as  the 
required  root,  and  expand  the  result  by  the  Binomial  Theorem. 

EXERCISE  119 

Find  the  approximate  values  of  the  following  to  five  decimal 
places : 

1.   VTT.     2.   V51.     3.   </60.     4.   ^/li.      5.   ^84.      6.   \/35. 


XXII.   PERMUTATIONS   AND    COMBINATIONS 

222.  The  letters  a  and  b  may  be  arranged  thus :  ab  or  ba. 
The  letters  a,  b,  and  c  may  be  arranged  two  at  a  time  thus  : 

ab,  ba,  ac,  ca,  be,  and  cb. 

The  different  orders  in  which  things  can  be  arranged  are 
called  their  Permutations. 

223.  General  Principle.  If  a  certain  thing  can  be  done  in  m 
different  ways,  and,  after  it  is  done,  if  a  second  thing  can  be 
done  in  n  different  ways,  then  the  two  things  can  be  done  in 
order  in  mn  different  ways. 

Example  1.  One  may  go  from  a  certain  city  to  another  by  two  dif- 
ferent railroads,  and  can  go  from  the  second  city  to  a  third  by  any  one  of 
three  different  railroads  Hence  one  can  go  from  the  first  to  the  third 
city  by  2  x  3  or  6  different  routes. 

(Having  made  the  first  part  of  the  trip  in  one  of  the  two  ways,  the 
trip  can  be  completed  in  any  one  of  three  ways,  making  three  different 
complete  routes ;  similarly  for  the  second  way  of  making  the  first  part  of 
the  trip.     This  makes  altogether  the  six  different  complete  routes.) 

224.  The  Permutations  of  n  Different  Things  Two  at  a  Time. 
Consider  the  n  letters  a,  6,  c,  •••.  These  are*  to  be  used  to  fill 
two  places,  a  first  place  and  a  second  place ;  as  ca,  where  c 
occupies  the  first  place  and  a  the  second  place. 

The  first  place  can  be  filled  by  any  one  of  the  n  letters; 
hence,  in  n  different  ways.  Then,  having  filled  the  first  place, 
the  second  place  can  be  filled  by  any  one  of  the  remaining 
(n  —  1)  letters ;  hence,  in  (n  —  1)  different  ways. 

Hence,  the  two  places  can  be  filled  in  n(n  —  1)  different 
ways,  according  to  the  principle  of  paragraph  223. 

225.  The  Permutations  of  n  Different  Things  r  at  a  Time. 
Consider  again  the  n  letters  a,  b,  c,  •••.  These  are  to  be  used 
to  fill  r  different  places. 

231 


232  ALGEBRA 

The  first  place  can  be  tilled  by  any  one  of  the  n  letters ; 
hence,  in  n  different  ways. 

The  second  place  can  then  be  filled  by  any  one  of  the  re- 
maining (n  —  1)  letters  ;  hence,  in  (w  —  1)  different  ways. 

Similarly  the  third  place  can  be  filled  in  (n  —  2)  different 
ways. 

Finally  the  rth  place  can  be  filled  in  n  —  (r  —  1)  or  (n— r-f-1) 
different  ways. 

Then,  according  to  the  general  principle  of  paragraph  223, 
the  whole  number  of  the  permutations  of  the  n  letters  taken 
r  at  a  time  is : 

n{n  -  l)(n-  2)(w  -  3)  ...  [*  -  r+  1). 

The  number  of  permutations  of  n  things  taken  r  at  a  time  is 
denoted  by  nPr.     Hence, 

nPr  =  n(n-l)(n-2)(n-3)  ...  (w_r  +  l). 

Note.  The  product  consists  of  factors  starting  with  the  number  n 
and  decreasing  by  1  each  time  until  the  number  of  factors  is  r. 

Example.  How  many  numbers  of  three  figures  each  can  be 
made  by  using  the  nine  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  no  digit 
is  used  twice  in  the  same  number  ? 

Solution  :  Each  arrangement  of  the  nine  digits  three  at  a  time  will  be 
a  different  number.  Hence,  the  whole  number  of  numbers  which  can  be 
formed  is  9P3,  or  9  .  8  .  7  ;  that  is,  504. 

226.  Clearly  nPn  =  n(n  -  l)(n  -  2)(ii  -3)  •«•  (» .—  »  +  1),  for 
r  =  ». 

Hence  nPn  =  n(n  -  l)(n  -  2)0  -  3)  ••  •  3  •  2  •  1. 

The  product  1  •  2  •  3  •••  (n  —  l)(w)  is  denoted  by  the  symbol 
n !,  and  is  called  "factorial  n." 

Hence  the  number  of  permutations  of  n  different  things 
taken  wata  time  is  n  factorial. 

Example.  The  permutations  a,  6,  and  c,  taken  all  at  a  time  is  3  •  2  ■  1 
or  6.     The  permutations  are  abc,  acb,  bac,  bca,  cab,  and  cba. 


PERMUTATIONS   AND   COMBINATIONS  233 

EXERCISE   120 

1.  How  many  permutations  can  be  formed  with  14  letters, 
taken  4  at  a  time  ? 

2.  In  how  many  different  orders  can  the  letters  of  the  word 
triangle  be  written,  taken  altogether  ? 

3.  A  certain  play  has  five  parts,  to  be  taken  by  a  company 
consisting  of  12  persons.  In  how  many  different  ways  can 
they  be  assigned  ? 

4.  How  many  different  numbers  of  4  different  figures  each 
can  be  formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9  if  no  digit 
occurs  twice  in  the  same  number  ? 

5.  Solve  the  example  formed  by  adding  to  the  statement  of 
Example  4  the  words  "and  if  each  number  is  to  begin  with  1." 

6.  How  many  different  numbers  of  6  different  figures  can 
be  formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  each  num- 
ber is  to  begin  with  2  and  is  to  end  with  9  ? 

7.  How  many  of  the  numbers  found  in  Example  6  have  the 
digit  5  as  one  of  their  digits  ? 

8.  How  many  even  numbers  of  five  different  figures  each 
can  be  formed  from  the  digits  4,  5,  6,  7,  8  ? 

9.  How  many  different  words  of  8  letters  each  can  be 
formed  from  the  letters  of  the  word  'ploughed,  if  the  third 
letter  must  be  o,  the  fourth  u,  and  the  seventh  e  ? 

10.  In  how  many  ways  can  a  teacher  arrange  6  boys  in  the 
6  front  seats  of  a  class  room  ? 

227.  The  Permutations  of  n  Things  taken  all  at  a  Time  if  the 
Things  are  not  all  Different. 

Special  Case.  Consider  the  distinguishable  permutations  of 
a,  a,  and  6,  taken  all  at  a  time.  They  are  aab,  aba,  baa.  If 
now  the  two  a's  are  replaced  by  ax  and  a2  respectively,  the 


234  .  ALGEBRA 

distinguishable  permutations  are :  axa2b,  a2a1b,  axba2,  a2baly 
baxa2,  and  ba2ax.  From  each  of  the  original  permutations,  two 
new  permutations  are  obtained.  The  result  is  the  same  as  the 
permutations  of  3  letters,  all  taken  together. 

General  Case.  Let  there  be  n  letters,  of  which  p  are  a's,  q 
are  6's,  and  r  are  c's,  the  rest  being  all  different.  Let  N  be 
the  number  of  different  permutations  of  these  letters  taken 
all  together. 

Suppose  that,  in  any  particular  permutation  of  the  n  letters, 
the  p  a's  are  replaced  by  p  new  letters  all  different,  and  differ- 
ing also  from  the  remaining  letters.  Then  by  permuting  these 
p  letters  in  all  possible  ways,  without  changing  the  positions 
of  the  remaining  letters,  p !  permutations  are  formed  from  the 
original  particular  permutation.     (§  226.) 

If  this  is  done  in  the  case  of  each  of  the  N  original  permuta- 
tions, the  whole  number  of  permutations  will  be  Nxpl. 

Again,  if,  in  any  one  of  these  JVxp !  permutations,  the  q  p's 
are  replaced  by  q  letters  all  differing  from  each  other  and  dif- 
fering also  from  all  of  the  remaining  letters,  then  by  permut- 
ing the  q  6's  in  all  possible  ways,  without  changing  the  order 
of  the  remaining  letters,  q !  permutations  are  formed  from  the 
original  permutation.  If  this  is  done  in  the  case  of  each  of 
the  Nxpl  permutations,  the  whole  number  of  permutations 
resulting  is  Nxpl  X  q\. 

In  like  manner,  if,  in  each  of  the  JVxpl  X  q\  permutations 
the  r  c's  are  replaced  by  r  new  letters,  all  different  and  differ- 
ing from  the  remaining  n  letters,  then,  by  permuting  them  in 
all  possible  ways,  r !  new  permutations  are  formed  from  each. 
The  total  number  of  permutations  is  N xp\  X  q\  X  r\. 

The  original  n  letters  have  now  been  replaced  by  n  letters 
all  different.  The  resulting  permutations  are  the  permutations 
of  n  different  letters  n  at  a  time ;  this  is  n\. 

Hence  N  X  p !  x  q !  x  r !  =  n !, 

or  N=- ■ -. 

p\  X  q\  X  r\ 


PERMUTATIONS   AND   COMBINATIONS  235 

Example.     How  many  permutations  can  be  made  from  the 
letters  in  the  word  Tennessee,  taken  all  together? 

Solution  :  1.   There  are  4  e's,  2  w's,  2  s's,  and  1 1. 

2.     ,^=       91       =l-2-8-4.5.6.7.8.9  =  6.fl.7.2<9  =  878k 
4!2!2!         1-2. 3- 4- 1-2. 12 


EXERCISE   121 

1.  In  how  many  different  orders  can  the  letters  of  the  word 
denomination  be  written  ? 

2.  There  are  4  white  billiard  balls  exactly  alike,  and  3  red 
balls,  also  alike.  In  how  many  different  orders  can  they  be 
arranged  ? 

3.  In  how  many  different  orders  can  the  letters  of  the  word 
independence  be  written  ? 

4.  How  many  different  signals  can  be  made  with  7  flags,  of 
which  2  are  blue,  3  red,  and  2  white,  if  all  are  hoisted  for  each 
signal  ?    • 

5.  How  many  different  numbers  of  8  digits  can  be  formed 
from  the  digits  4,  4,  3,  3,  3,  2,  2,  1? 

228.  The  Combinations  of  things  are  the  different  collections 
which  can  be  formed  from  them  without  regard  to  the  order 
in  which  they  are  placed. 

Thus,  the  combinations  of  the  letters  a,  6,  c,  taken  two  at  a  time,  are 
ab,  be,  ca  ;  for'  though  ab  and  ba  are  different  permutations,  they  form 
the  same  combination. 

The  number  of  combinations  of  n  different  things  taken  rata  time  is 
usually  denoted  by  the  symbol  nCr. 

229.  The  number  of  combinations  of  n  different  things  taken  r 
at  a  time. 

The  number  of  permutations  of  n  different  things  taken  r  at 
a  time  is       ^  _  1)(n_2)  ...  (n  _  r  +  1)  (§  225). 


236  ALGEBRA 

But,  by  §  226,  each  combination  of  r  different  things  may- 
have  r !  permutations. 

Hence,  the  number  of  combinations  of  n  different  things  taken 
rata  time  equals  the  number  of  permutations  divided  by  ?*!. 

Thatis,       >c,f  =  n(»-l)(n-2)...(»-r  +  l).  (3) 

T  ! 

230.  Multiply  both  terms  of  the  fraction  (3)  by  the  product 
of  the  natural  numbers  from  1  to  n  —  r  inclusive ;  then 

p  _  n(n  —  1)  •  •  •  (n  —  r  4- 1)  •  (n  —  r)  -  •  2  •  1  _        n ! 
n   r~  r!  xl  .2..."(n"-r)  ~r\(n-r)\' 

which  is  another  form  of  the  result. 

231.  The  number  of  combinations  of  n  different  things  taken  r 
at  a  time  equals  the  number  of  combinations  taken  n  —  r  at  a  time. 

For,  for  every  selection  of  r  things  out  of  n,  we  leave  a  selec- 
tion of  n  —  r  things. 

The  theorem  may  also  be  proved  by  substituting  n  —  r  for  r,  in  the 
result  of  §  230. 

Example  1.  How  many  different  combinations  can  be 
formed  with  16  letters,  taking  12  at  a  time  ? 

Solution  :  By  §  231,  the  number  of  combinations  of  16  different  things, 
taken  12  at  a  time,  equals  the  number  of  combinations  of  16  different 
things,  taken  4  at  a  time. 

Putting  n  =  16,  r  =  4,  in  (3),  §  229, 

p  _16.15.14.13,qO0 
16°4"      1.2.3-4      -1820, 

Example  2.  How  many  different  words,  each  consisting  of 
4  consonants  and  2  vowels,  can  be  formed  from  8  consonants 
and  4  vowels  ? 

The  number  of  combinations  of  the  8  consonants,  taken  4  at  a  time,  is 

8    7-6-5,or70. 
1-2-3-4' 


PERMUTATIONS   AND   COMBINATIONS  237 

The  number  of  combinations  of  the  4  vowels,  taken  2  at  a  time,  is 

i^,or6. 
1-2' 

Any  one  of  the  70  sets  of  consonants  may  be  associated  with  any  one 
of  the  6  sets  of  vowels  ;  hence,  there  are  in  all  70  x  6,  or  420  sets,  each 
containing  4  consonants  and  2  vowels. 

But  each  set  of  6  letters  may  have  6 !,  or  720  different  permutations 
(§  226). 

Therefore,  the  whole  number  of  different  words  is 

420  x  720,  or  302400. 


EXERCISE   122 

1.  How  many  combinations  can  be  formed  from  15  things, 
taken  5  at  a  time  ? 

2.  How  many  combinations  can  be  formed  from  17  things, 
taken  11  at  a  time  ? 

3.  How  many  different  committees,  of  8  persons  each,  can 
be  selected  from  14  persons  ? 

4.  There  are  5  points  in  a  plane,  no  3  being  in  'the  same 
straight  line.  How  many  straight  lines  are  determined  by 
them? 

5.  How  many  different  words,  each  having  5  consonants 
and  1  vowel,  can  be  formed  from  13  consonants  and  4  vowels  ? 

EXERCISE  123 
Miscellaneous  Examples 

1.  There  are  11  points  in  a  plane,  no  3  in  the  same  straight 
line.  How  many  different  quadrilaterals  can  be  formed,  hav- 
ing 4  of  the  points  for  vertices  ? 

2.  From  a  pack  of  52  cards,  how  many  different  hands 
of  6  cards  each  can  be  dealt  ? 

3.  How  many  different  numbers  of  7  figures  each  can  be 
formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  the  first, 
fourth,  and  last  digits  must  be  odd  numbers  ? 


238  ALGEBRA 

4.  Out  of  10  soldiers  and  15  sailors,  how  many  different 
parties  can  be  formed,  each  consisting  of  3  soldiers  and  3 
sailors  ? 

5.  Out  of  3  capitals,  6  consonants,  and  4  vowels,  how  many 
different  words  of  6  letters  each  can  be  formed,  each  beginning 
with  a  capital,  and  having  3  consonants  and  2  vowels  ? 

6.  How  many  points  of  intersection  are  determined  by  6 
straight  lines  if  no  3  of  the  lines  pass  through  the  same  point, 
and  if  no  2  are  parallel  ? 

7.  How  many  different  words  of  8  letters  each  can  be 
formed  from  8  letters,  if  4  of  the  letters  cannot  be  separated  ? 

8.  In  how  many  ways  can  a  committee  of  2  teachers  and 
3  students  be  selected  from  5  teachers  and  10  students  ? 

9.  In  how  many  different  ways  may  10  students  be  seated 
in  15  seats  ?     (Leave  result  in  factored  form.) 

10.  How  many  games  will*  be  played  in  a  baseball  league 
of  8  teams  if  each  team  plays  10  games  with  each  of  the  other 
teams  ? 

11.  How  many  signals  can  be  made  with  1  red,  1  white, 
and  1  blue  flag,  using  them  either  1  at  a  time,  2  at  a  time, 
or  all  together,  if  the  order  in  which  the  flags  are  shown  con- 
stitutes a  part  of  the  signal  ? 

12.  In  how  many  different  ways  can  a  captain  of  a  baseball 
team  arrange  his  batting  list  of  9  men  if  he  wishes  certain 
3  men  to  bat  in  the  order  1,  4,  and  7  ? 

13.  How  many  different  multiplication  facts  are  involved 
in  the  multiplication  table  from  lxl  up  to  9x9,  if  (for 
example)  5x4  and  4x5  are  considered  one  fact  ? 


XXIII.   DETERMINANTS 


232.   The  symbol 


3     4 

2     7 


is  called  a  determinant.     Its  value 


is  defined  to  be  3  •  7  —  2  •  4,  which,  equals  21  —  8,  or  13. 
a    c 


In  general 

b     d 

and  is  defined  thus : 


is  called  a  Determinant  of  the  Second  Order 


a    c 
b    d 


=  ad  —  be. 


The  numbers  a,  b,  c,  and  d  are  called  the  elements  of  the  de- 
terminant. 

Clearly,  any  difference  such  as  rs  —  mn  may  be  arranged  as 

r    n 
a  determinant :  thus  rs  —  mn 

m    s 


Example  1. 


Example  2. 


=  2-3-  4(-  5)  =  6  +  20  =  26. 

L5  =  2- 13-3.  5=    2      5|. 
3    13 


EXERCISE  124 


Find  the  values  of : 


2. 


6    5 

•       3. 

4     -2 

•      5. 

-5     3 

•       7. 

2m    —  p 

4    2 

6         9 

2     6 

2n         r 

5      3 

2  -7 

•     4. 

3  -4 

-2      7 

•     6. 

3a    4 
2c     1 

8. 

Sa     Id 

2  c     5e  ' 

Express  as  determinants : 
9.   mn  —  xy.  11.   33  —  14. 

10.   2ab      cd.  12.    6c  —  5  d. 

239 


13.  cd  +pq. 

14.  3  mn  +  2  rs. 


240 


ALGEBRA 


233.  Determinants  make  it  possible  to  solve  simultaneous 
linear  equations  by  inspection.  Solving  the  following  pair  of 
equations, 

ax  +  by  =  c 
clx+  ey=f 


ce  —  bf 
ae  —  bd 


H  and  y  = 


af—  cd 

ae  —  bd 


x  = 


c    b 

a     c 

f    e 

and  y  = 

d    f 

a     b 

a     b 

d    e 

d     e 

Notice  that  the  two  solutions  may  be  expressed  as  the  quo- 
tients of  determinants  whose  terms  are  the  coefficients  of  the 
equations. 

Rule.  —  To  solve  two  simultaneous  linear  equations  having  two 
unknowns  by  determinants : 

1.   Arrange  the  equations  in  the  form  :      J  '  ~    ' 

\dx  -\-ey  =  f. 

its  denominator  is  the  deter- 
a     b 
d    e 


2.    The  value  of  x  is  a  fraction 
minant  formed  by  the  coefficients  of  x  and  / 


its  numera- 


tor is  the  determinant  obtained  by  replacing  the  coefficients  of  x  in 
the  denominator  determinant  by  the  corresponding  absolute  terms, 

c    b 

f    e 

3.  The  value  of  /  is  a  fraction  with  the  same  denominator  as  x ; 
its  numerator  is  the  determinant  obtained  by  replacing  the  coeffi- 
cients of  /  in  the  denominator  determinant  by  the  absolute  terms, 

a     c 

d     f 

Example.     Solve  the  pair  of  equations  :   J  ^         „ 


16. 


DETERMINANTS 


241 


Solution  :  x  = 


-16 

-5 

7 

1        2 

-5 

3 

7 

_  -i6-7-5(-5) 
2-7-3(-5) 


■112  +  25  _  -87 


14  +  15 


29 


-3. 


2      -16 1 

3 jjj  =  10- 3(- 16)  ^10+48^58^ 

2        Z5]     2.7-3(-6)      14  +  15      29  ~    ' 


Check:   In  (1) :  Does  2(- 3)- 5(2)  =- 16  ?     Does  -  6  -  10  =-16? 
Yes.     In  (2):  Does  3(- 3) +7(2)  =  6?     Does -9 +  14  =  5?    Yes. 


EXERCISE   125 

Solve  the  following  equations  by  determinants : 


1. 


2. 


4. 


f6a  +  5?/  =  28. 
J4#+    y  =  14. 

f      7  x  —  9  ?/  =  15. 
|-5oj  +  8y  =  -17. 

f5^-67/  =  -9. 
(3a;-  5  ?/  =  — 4. 


8  m  - 15  v  =  18. 
12  m+   6v  =  -ll. 


5. 


6. 


7. 


[5p  + 

2r  = 

-4. 

Ui>- 

11  r  = 

-45 

j4r- 

3*  =  - 

18. 

5s=- 

7. 

f3«- 

4^/  =  - 

11. 

^  +  5 

5 

2/  +  1 

-0. 

|  ma?  - 

■  ???/  as  mn. 

\  m'sc  +-  n'y  =  m  V. 

234.    Determinants  are  especially  useful  in  solving  simul- 
taneous linear  equations  with  more  than  two  unknowns. 

<*1        «2        «3 

bx     b2     63 
C\     c2      c3 

is  called  a  determinant  of  the  third  order.     Its  value  is  defined 
to  be: 

&1&2C3  +  «2^3C1  +  «3C2&1  —  C!&2«3  ~  &1^2C3  —  UfilPl- 


242 


ALGEBRA 


The  adjoining  diagram  aids  in  recalling  this 
value.  Take  the  product  aiboCs  along  the  di- 
agonal and  add  to  it  the  two  products  formed 
by  starting  with  a2  and  a3  respectively  and  fol- 
lowing the  arrows  which  point  in  the  direction 
of  this  diagonal ;  then  subtract  the  product 
ci&2«3  along  the  other  diagonal,  and  also  sub- 
tract the  two  other  products  formed  by  starting 
with  b\  and  a,i  respectively  and  following  the 
arrows  which  point  in  the  direction  of  this 
second  diagonal. 


Example. 


=  1. 7.6  +  5-3. 2  +  2(-3). 4 

-2.7.2-4.5.6-I.3.  (-3) 
=  42+30-24-28-120+9 
=  -91. 


Find  the  values  of : 


EXERCISE  126 


1     2    3 

2 

4     6 

2       2 

3 

2    12 

2. 

3  - 

-2    3 

'3. 

-2  -4 

-11 

3    3    1 

1 

5    4 

5  -6 

2 

4.    Solve  the  equations  : 


3  x  +  y  —  z  — 14. 
x  +  3  y  —  z  =  16. 
x  +  y  —  3  z  =  — 10. 


Solution  :    A  rule  similar  to  that  of  §  345  applies  for  linear  equations 
with  more  than  two  unknowns.     Hence : 


14  1  -  1 
10  3  -1 
10    1     -3 


3  1  -1 
13-1 
1     1     -3 


_  126  +  10  -  16  -  30  +  48  +  14  _  -  100  _  g 
_  27 -1-1+3+3+3  -20 


DETERMINANTS 


243 


3 

14 

-1 

1 

16 

-1 

1 

-10 

-3 

-  120 

3 

1 

-  1 

-20 

1 

3 

-1 

1 

1 

-3 

=6. 


3 

1 

14 

1 

8 

16 

1 

1 

-10 

3 

1 

-  1 

1 

3 

-1 

1 

1 

-3 

140 


-20 


=  7. 


Check  :     The  solution  checks  when  substituted  in  the  three  equations. 

Note.     The  equations  must  be  arranged  first  in  the  form  ax+by+cz  =  d. 
Thus  the  equation  2  x  —  .3  z  =  7  would  be  written  2x  -\-  0  y  —  S  z  =  1. 


Solve  the  following  equations  by  determinants : 


5. 


x  +  y  —  z  =  24. 
4  x  -4-  3  y  —  x  =  61. 
6x—  5y—     z  ™  11. 


5a!—    y  +  4  z  =  —  5. 

6.     , 

3x  +  5y+  6z  =  -20. 

x  +  3y-$z  =  -27. 

Aa-5b-6c=     22. 

7-  1 

a—     b  +     c  =  —    6. 

9  a           +     c=      12. 

8. 


9. 


4  x 

-3*/  = 

1. 

4,y 

-3z  = 

-15. 

4z 

-3a?  = 

10. 

9a; 

+  5z  = 

-7. 

3a? 

+  -5.y  = 

:1. 

9y 

+  3z  = 

2. 

f  2#  +  5?/  +  3z  =  -7. 
10-       2?/-4z  =  2-3a\ 
[5a;  +  92/  =  5-f-7z. 


DETERMINANTS  OF  ANY  ORDER 

235.  If  the  numbers  1,  2,  3,  4,  5,  •••  n  are  arranged  in  any 
other  order,  each  instance  when  a  greater  number  precedes  a 
less  is  called  an  Inversion. 

Thus,  for  the  numbers  1,  2,  3,  4,  5,  the  arrangement  51432  has  7  inver- 
sions :  5  before  1,  before  2,  before  3,  and  before  4  ;  4  before  3,  and  before 
2  ;  and  3  before  2. 

(Ill        Q>\2       0tI3       •••       0£ln 
a2i        d'22       #23        '"        a2n 


236.   The  symbol 


#nl       <*«2        «n3 


,  having  n  rows  of 


244  ALGEBRA 

elements,  each  row  consisting  of  n  elements  is  a  Determinant 
of  the  nth  Order. 

Note  1.  The  first  number  of  the  subscript  of  an  element  denotes  the 
row  in  which  the  element  lies,  and  the  second  denotes  the  column.  Thus, 
a35,  read  "  a-three-five,"  is  in  the  third  row  and  fifth  column. 

237.    Definition  of  the  Value  of  a  Determinant. 

1.  Form  all  possible  products  of  the  elements  of  the  deter- 
minant, such  that  each  product  shall  have  as  factors  one  and 
only  one  element  from  each  row,  and  one  and  only  one  from 
each  column. 

2.  Arrange  the  elements  in  each  product  so  that  the  first 
subscripts  are  in  the  order  1,  2,  3,  •••  n. 

3.  Make  the  product  positive  or  negative  according  as  the 
number  of  inversions  in  the  second  subscripts  is  even  or  odd. 


Thus, 


an 

«12 

«13 

«21 

«22 

«23 

«31 

«32 

«33 

=  <2ll#22#33  —    «11«23«32   —    «12«21«33    +  «12«23«31 
+  ^13^21^32  —  «13#22#31^ 


Note  1.     This  value  agrees  with  that  found  as  in  §  234. 

Note  2.  The  elements  lying  in  the  diagonal  joining  the  upper  left 
hand  element  with  the  lower  right  hand  element  form  the  principal  diag- 
onal.    The  product  of  these  elements  is  always  positive. 

238.  Consider  any  term  of  a  determinant  of  the  fourth 
order,  as  aua2lazzaA2.  The  number  of  inversions  in  the  second 
subscripts  is  4,  an  even  number.  Consider  any  other  arrange- 
ment of  these  factors  ;  as,  azza^aua.2v  The  total  number  of  in- 
versions among  the  first  and  the  second  subscripts  is  8,  again  an 
even  number. 

In  general,  if  the  number  of  inversions  for  any  arrangement 
of  the  elements  in  a  term  of  an  expanded  determinant  is  even,  the 
number  of  inversions  remains  even  for  any  other  arrangement  of 
the  elements  in  that  term;  similarly,  if  the  number  of  inversions 
is  odd,  it  remains  odd. 


DETERMINANTS  245 

(a)  As  a  consequence  of  this  fact,  the  expansion  of  a  deter- 
minant may  have  the  elements  of  each  product  arranged  so 
that  the  second  subscripts  are  in  the  order,  1,  2,  3,  •••,  n,  giving 
each  product  the  plus  or  minus  sign  according  as  the  number 
of  inversions  in  the  first  subscripts  is  even  or  odd. 

Thus, 

«11«22«33  —  «11«23«32  —  «12«521«33  +  #12#23#31  +  «13«21«32  —  ai3#22«31 

may  be  written 

«11«22«33  —  «11«32«23  —  «21«12«33  +  «31«12«23  +  «21«32«13  ~  «31«22«13- 

Examination  will  show  that  the  signs  are  correct. 

(b)  A  second  immediate  consequence  is  that  the  elements 
of  each  product  may  be  arranged  in  any  manner,  provided  the 
sign  of  each  term  is  determined  by  the  total  number  of  inver- 
sions among  both  the  first  subscripts  and  the  second  subscripts 
of  the  term. 

PROPERTIES   OF  A  DETERMINANT 

239.  A  determinant  is  not  altered  in  value  if  its  rows  are 
changed  to  columns,  and  its  columns  to  rows. 

Thus,  it  will  be  proved  that 


an 

ai2 

«13 

a2i 

a22 

a23 

= 

a3i 

a32 

«33 

an 

a2i 

a3i 

ai2 

«22 

a32 

ai3 

a23 

a33 

Proof.  The  second  subscripts  of  the  first  determinant  are 
the  same  as  the  first  subscripts  of  the  second  determinant. 
The  number  of  rows  and  columns  in  each  is  the  same.  If 
the  first  determinant  is  expanded  by  the  rule  in  §  237  and 
the  second  by  the  remark  (a)  in  §  238,  the  results  are  the 
same. 

Thus,  the  determinants  are  respectively  : 

aii«22a33  —  ana23a32  +  ai2«23a3i  —  ai2a2i«33  +  ai3a2ia32  —  ai3«22a3i 


246 


ALGEBRA 


and 

«11«22«33  —  «11«32«23  _  «21«12«33  +  «21«32«13  +  #31#12#23  —  «31«22«13- 

These  are  equal  except  for  the  arrangement  of  the  elements  in  the 
terms. 

240.    A  determinant  is  changed  in  sign  if  any  two  consecu- 
tive rows,  or  any  two  consecutive  columns,  are  interchanged. 

Thus,  it  will  be  proved  that 


an 

«12 

«13 

an 

«12 

ai3 

«21 

#22 

«23 

=  - 

a3i 

«32 

^33 

«31 

«32 

«33 

«21 

a22 

a23 

Proof.  Consider  any  term  of  the  first  determinant;  as, 
a12a2ia33.  The  sign  of  this  term  in  the  first  determinant  is 
minus,  as  there  is  one  inversion. 

This  same  term  occurs  in  the  expansion  of  the  second  de- 
terminant, as  the  term  has  one  and  only  one  element  from  each 
row  and  each  column,  and  the  rows  and  columns  of  the  second 
determinant  are  the  same,  except  for  order,  as  those  of  the 
first  determinant.  In  fact,  this  term  is  the  product  of  the  ele- 
ments printed  in  black  type. 

Oi2a2ia33)  considered  a  term  of  the  second  determinant  ex- 
panded according  to  the  remark  (p)  of  §  238,  has  the  same 
inversions  among  its  second  subscripts  (213)  as  when  it  is 
considered  a  term  of  the  first  determinant.  Its  first  sub- 
scripts (123)  show  one  inversion,  namely  2  before  8,  as  the 
proper  order  of  the  rows  in  the  second  determinant  is  132. 
Hence  there  is  one  more  inversion  among  the  subscripts  in 
this  case,  making  two,  and  hence  the  sign  is  plus. 

In  a  similar  manner,  it  may  be  proved  that  every  term  of 
each  of  the  determinants  is  also  a  term  of  the  other  determi- 
nant if  the  sign  of  the  term  is  changed.  This  proves  the  the- 
orem stated. 

241.  A  determinant  is  changed  in  sign  if  any  two  rows,  or 
any  two  columns,  are  interchanged. 


DETERMINANTS 


247 


Thus,  it  will  be  proved  that 


an 

«12 

an 

«21 

a22 

«23 

=  — 

«31 

«32 

«33 

«31 

«32 

«33 

«21 

«22 

«23 

an 

«12 

«13 

Proof.     To  change  the  first  determinant  into  the  second, 
interchange  the  first  row  and  second  row  of  the  first  determi- 

#2i       #22 


nant,  getting  the  determinant 


«, 


,  and  then   inter- 


change the  second  row  of  this  new  determinant  with  the  third 
row,  getting 


21        #22        #2 
31        #32        #3 

11 


Now   interchange   the   first   and 

#11        #12        #13 

second  rows  of  the  last  determinant,  and  the  desired  determi- 


nant 


#31        «32 
#21        #22 


#33 


#1 


a, 


is  obtained. 


All  together  three  interchanges  of  consecutive  rows  have  been 
made.  Each  causes  a  change  in  the  sign  of  the  determinant. 
The  resulting  determinant  is  the  negative  of  the  given  deter- 
minant. 

A  similar  proof  may  be  given  for  a  determinant  of  any 
order. 

242.  Cyclical  Interchange  of  Rows  or  Columns.  It  will  be 
proved  that 


#22 


=    (-1) 


(n-1) 


#21 

#22 

"•    «2» 

#31 

#32 

'••    #3„ 

«»1 

«n2 

••  ann 

#11 

#12 

...  #ln 

Proof.     The  first  row  has  been  made  to  occupy  the  position 
of  the  nth  row.     This  may  be  accomplished  by  interchanging 


248 


ALGEBRA 


the  first  row  with  the  second,  then  with  the  third,  and  so  on 
up  to  the  nth  inclusive.  This  makes  all  together  (n  —  1)  inter- 
changes of  adjacent  rows,  and  causes  (n  —  1)  changes  in  sign  of 
the  determinant.     Hence  the  fact  stated  above  is  true. 

243.  If  two  rows,  or  two  columns,  of  a  determinant  are 
identical,  the  value  of  the  determinant  is  zero. 

Proof.  Let  D  be  the  value  of  the  original  determinant  hav- 
ing two  rows  identical. 

If  these  two  rows  are  interchanged,  the  value  of  the  result- 
ing determinant  is  —  D  (§  241).  But  the  two  determinants 
are  actually  identical,  since  the  rows  that  were  interchanged 
are  identical. 

Hence         D  =  -  D,  or  2  D  =  0.     Therefore  D  =  0. 

244.  If  each  element  of  one  row,  or  of  one  column,  is  a  bi- 
nomial, the  determinant  can  be  expressed  as  the  sum  of  two 
determinants. 

Thus,  it  will  be  proved  that 


1>1  + Ci       «12       «13 

62  +  C2       «22       «23 

63  +  C3       «32       «33 


h 

«12 

«13 

1>2 

^22 

«23 

+ 

h. 

«32 

033 

<*4 

«12 

«13 

di 

«22 

«23 

cs 

«23 

«33 

Proof.  Each  term  of  the  first  determinant  is  the  product 
of  a  binomial  from  column  one  and  a  monomial  from  each  of 
the  other  columns. 

Consider  (bl  +  cl)a22a33.  This  equals  bla22azz-\-cla22a3S.  The 
result  is  obviously  the  sum  of  the  two  corresponding  terms  of 
the  other  two  determinants. 

In  a  similar  manner,  it  may  be  proved  that  each  term  of  the 
first  determinant  is  the  sum  of  the  two  corresponding  terms  of 
the  other  two  determinants.  Hence,  the  first  determinant  is 
the  sum  of  the  other  two. 


Note.     If  any  column,  or  any  row,  consists  of  the  sum  of  n  terms, 
then  the  determinant  may  be  expressed  as  the  sum  of  n  determinants. 


DETERMINANTS 


249 


245.  If  all  of  the  elements  in  one  row,  or  in  one  column, 
are  multiplied  by  the  same  number,  the  determinant  is  multi- 
plied by  that  number. 

Thus,  it  will  be  proved  that 


anr 

Op 

#13 

a2\r 

#22 

«23 

=  r 

azir 

#32 

#33 

an 

#12 

#13 

#21 

#22 

«23 

«31 

«32 

a33 

Proof.  Since  each  term  of  the  expanded  determinant  on 
the  left  contains  one  and  only  one  factor  from  the  first  column, 
each  term  must  have  one  and  only  one  factor  r.  Hence  r  is  a 
common  factor  of  the  terms  of  the  determinant.  This  proves 
the  theorem. 

246.  If  all  of  the  elements  of  one  column,  or  of  one  row,  of  a 
determinant  be  multiplied  by  the  same  number,  and  either  added 
to  or  subtracted  from  the  corresponding  elements  of  another 
column,  or  row,  the  value  of  the  determinant  is  not  changed. 

Thus,  it  will  be  proved  that 


#ii+&#u 

#12    ai3 

On 

#12 

#13 

«21  +  ##23       #22      #23 

= 

#21 

#22       #23 

#31  +  &a33       #32       #33 

#81 

#32       #33 

Proof. 

Oa+kOu 

#12        #13 

#11        #12 

«M 

k,an     a12     a13 

a2l  +  ka23 

#22        #23 

= 

#21        #22 

"'J:! 

+ 

23        #2^           2* 

#31  4"  ^#33 

#32       #33 

#31        #32 

":a 

^#33        #32        #33 

#11        #12 

«13 

#13        #12        #13 

= 

#21        #22 

hz 

+  k 

#23        #22        #23 

#31        # 

32 

'-.a 

#33 

#32        #33 

#12       #13 


#21        #22 


+  k  •  0. 


(§  244) 


(§  245) 


(§  243) 


This  clearly  proves  the  theorem. 


250 


ALGEBRA 


Thus,  in 


247.  Minors.  If  the  elements  of  the  row  and  of  the  column 
of  a  determinant,  in  which  any  particular  element  lies,  are 
omitted  from  the  determinant,  the  resulting  determinant  is 
called  the  Complementary  Minor  of  that  particular  element. 

an    <ti2    «i3 

mj^ — <^ — e^.   is  the  complementary  minor  of  a22- 

«31       '^32       #33 

For  brevity,  the  complementary  minor  of  a22  is  denoted  by 
A22 ;  in  general,  of  aik,  by  Aik. 

Note.  If  the  given  determinant  is  of  order  n.  the  complementary 
minor  of  one  of  its  terms  is  of  order  (n  —  1). 

248.  The  coefficient  of  an  in  the  expansion  of 


Oil 

ai2  • 

'  <hn 

«21 

«22  • 

'  <hn 

«W 

««2   ' 

-   <*nn 

4;  *.€. 


<x22     a23 

^'32       ^33 


«2n 

«3n 


n2       an3 

Proof.  The  absolute  value  of  the  terms  which  have  the 
element  an  as  a  factor  are  obtained  by  forming  in  all  possible 
ways  the  products  of  an  by  the  other  elements  of  the  deter- 
minant, subject  only  to  the  restriction  that  there  shall  be  one 
and  only  one  element  from  each  row  except  the  first,  and  one 
and  only  one  element  from  each  column  except  the  first. 
From  this,  it  is  evident  that,  except  possibly  for  their  signs, 
the  coefficient  of  an  may  be  obtained  by  forming  all  possible 
products  of  the  following  elements  taken  (n  —  1)  at  a  time, 

^22       ^23  *'*  ^2/i 
^32       ^33  '"  ®Zn 


<'n2 


subject  to  the  restriction  that  each  product  shall  have  one  and 
only  one  element  from  each  row  and  each  column. 

The  sign  of  any  term  of  the  original  determinant  containing 


DETERMINANTS 


251 


an,  is  determined  by  the  inversions  of  the  remaining  factors 
of  the  term,  if  the  term  an  appears  as  the  first  factor  of  the 
term.  (§  237.)  But  this  sign  will  be  exactly  the  sign  of  the 
corresponding  term  of  the  determinant  An.  Hence  the  coeffi- 
cient of  an  is  An. 

249.    Coefficient  of  Any  Element  of   a  Determinant    The  co- 
ot, 
efficient  of  a32  of  the  determinant 


ii 

«21 


an 

an 

«22 

«23 

&32 

«33 

is  (-  1)2+3A 


Proof.     1.    By  two  interchanges  of  consecutive  rows,  the 
last  row  of  the  given  determinant  may  be  made  the  first ;  then 


(§  240) 


2.   By  interchanging  the  first  two  columns  of  the  last  deter- 
minant, the  second  column  may  be  made  the  first ;  then, 


«11 

a12 

«13 

<% 

#32 

a33 

«21 

a22 

#23 

=   (- 

"I)2 

au 

«12 

«13 

«31 

a32 

«33 

a2l 

a22 

«23 

a„ 

a12 

«13 

«32 

«31 

«33 

«21 

#22 

CT-23 

=  (" 

■1)3 

«12 

an 

<*ll 

«31 

«32 

tt33 

#22 

«21 

a23 

(§  240) 


3.  Hence  the  expansion  of  the  first  determinant  may  be  ob- 
tained by  considering  the  expansion  of  the  second  determinant 
of  step  2. 

.-.  the  coefficient  of  a32  is  (—  l)3 


tt23 


(§  248) 


4.    But 


is  the  minor  Ai2  of  the  given  determinant. 


o„     a13 

a21     ot23 

Note  also  that  (—  l)3  is  the  same  as  (—  1)3+2. 
{(—  1)3+2  is  used  as  a  matter  of  convenience  in  this  case}. 
Then  the  coefficient  of  a32  =  (—  l)3+2^432. 
Note  that  the  exponent  of  (—  1)  is  the  sum  of  the  subscripts 
of  the  element. 


252 


ALGEBRA 


This  fact  is  a  special  case  of  the  general  theorem :   in   a 
determinant  of  the  ?*th  order,  the  coefficient  of  al;  is  (—  l)i+iAir 

Proof.     1.    By  (i  —  1)  interchanges  of  adjacent  rows,  the 
ith  row  may  be  made  the  first  row. 

2.  Then  the  original  determinant  D  =  (—  l)f_1Z>',  where  Dr 
is  the  new  determinant  obtained  in  step  1.  (§  240) 

3.  By   (j  —  1)  interchanges  of   adjacent   columns,   the  jth 
column  can  be  made  the  first. 

4.  Then  D  =  (—  l)*'-14'-1!)",  where  D"  is  the  new  determinant 
obtained  in  step  3.  (§  240) 


That  is 


an 

ai2  • 

•    Ctln 

«21 

a22  • 

•    «2n 

«nl 

«n2  • 

.  ann 

=  (-  iy+s-* 


<*H 

an 

aii  • 

.  *  . 

••  ain 

a\j 

an 

«i2  •• 

.  *  . 

•  au 

a»j 

0*1 

(*22   • 

* 

•   «2„ 

* 

* 

*      . 

.  *  . 

..    * 

ani 

an\ 

ani  • 

.  * 

ann 

Note.     The  *'s  indicate  the  places  formerly  occupied  by  the  ith  row 
and  jth  column. 


aiX 

«12 

...     *     . 

"•     ain 

a21 

(^22 

...     * 

••     «2n 

# 

* 

...     *     . 

.       * 

«nl 

«n2 

...     *     . 

'    *m 

5.   .-.  the  coefficient  of  cit.=  (-l)<+>'-2 

=  (-  lY*4r 
Note.     (-l)*+>  =  (-l)»'+>-2. 

250.   A  determinant  of  the  fourth  order  may  be  written : 


a,,    a,,     a. 


U        "-12       "13       "14 
GE21        ^22        ^23        ^24 

a31     of32     <z33    a34 


^41       ^42 


44 


=  anAn  +  (-  l)1+2ai2A2  +  (-  I)1+3«i3^i3 
=  anAn  -  anAn  +  anAn  -  auAu.     (§  249) 


DETERMINANTS 


253 


Notice  that  the  elements  of  the  first  row  are  multiplied  by 
their  respective  minors  and  that  the  products  have  alternately 
the  signs  +  and  — . 

Similar  expansions  may  be  given  for  determinants  of  any 
order. 

251.   Evaluation  of  Determinants.     The  theorems  proved  in 

§§  239  to  250  make   it   possible   to   shorten   the   process   of 

evaluating  a  determinant,  especially  of  order  higher  than  the 

third. 

5       7       8       6 

11     16     13     11 
Example  1.     Evaluate      1  .     ~.     ^     «o 

7     13     12      2 


Solution  :  1.  Subtracting  the  first  row  from  the  last,  twice  the  first 
row  from  the  second,  and  three  times  the  first  row  from  the  third,  the 
determinant  becomes  by  §  246, 


5 

7 

8 

6 

1 

2 

-3 

-  1 

=  2 

1 

3 

-4 

5 

2 

6 

4 

-4 

6  7  8  6 
12-3-1 
13-4         5 

13         2-2 


,  by  §  245. 


2.  Subtracting  five  times  the  second  row  from  the  first,  adding  the 
second  row  to  the  third,  and  subtracting  the  second  row  from  the  fourth, 
the  last  determinant  becomes 


0 

-3 

23        11 

1 

0 

2 
5 

-3     -1 

-7         4 

=  2   j  0  •  An  -  1  An  4 

0 

1 

5     -  1 

-  3       23       11 

=  -2J21 

=  -2 

5-7         4 
1         5-1 

+  0Ai 


oaA 


(§  250) 


The  object  of  all  these  changes  is  to  put  the  given  determinant  into 
such  form  that  all  but  one  of  the  elements  in  one  column  (or  one  row) 
are  zero. 


254 


ALGEBRA 


In  the  last  determinant,  subtract  five  times  the  first  column  from  the 
second,  and  add  the  first  column  to  the  last.    Then 


-2 


23       11 

-7         4 

5     -1 


=  -2 


3 

38 

8 

5 

-32 

9 

1 

0 

0 

=  -2       I- 


f    |-32 
-2(342  +  256)  =-1196. 


0  •  An  +  0  •  ^33 


Another  method  of  solution  is  illustrated  in 


Example  2.   Evaluate 


y    V 
z     z2 


Solution:  1.     If  x  is  set  equal  to  y,  two 
therefore  the  determinant  vanishes. 

Hence  (x  —  y)  is  a  factor  of  the  determinant. 

2.  Similarly  (y  —  z)  and  z  —  x  are  factors. 

3.  x  is  a  factor,  since  it  is  a  factor  of  the  first  row. 
Similarly  y  and  z  are  factors. 

4.  .-.  the  determinant  =  cxyz(x  —  y)(y  —  z)(z—  x). 
The  determinant  is  of  the  sixth  degree  in  a;,  y,  and  z. 
The  factors  found  give  an  expression  of  the  sixth  degree,  provided  c  is 

a  constant. 

5.  The  leading  term  is  xy2zz.     This  term  will  appear  in  the  product  of 
step  4,  if  c  sr  1. 

.  \  the  determinant  =  xyz(x  —  y)(y  —  z)  (z  —  x). 


rows  are  identical,  and 

(§243) 

(§94) 

(§  245) 
(§18) 


EXERCISE   127 
Evaluate  the  following : 


7      8      9 

25   23 

19 

Ill 

28    35    40 

3. 

14    11      9 

5. 

se2   y2   z2 

. 

21    26    30 

21    17    14 

a?   y3   z3 

9    13    17 

1    a   a3 

a    b    a  +  b 

11    15    19 

4. 

1    b    b3 

6. 

b    c    b  +  c 

17    21    25 

1    c    c3 

c    Q>    c-f 

■  a 

DETERMINANTS 


255 


7. 


1 

1 

1 

1 

6 

11 

25 

7 

8 

7 

1 

3 

1 

6 

6 

6 

13. 


a 

a 

b 

b 

b 

b 

b 

a 

a 

b 

a 

b 

8. 


3 

1 

5 

2 

4 

10 

14 

6 

8 

9 

1 

4 

B 

15 

21 

9 

14. 


0 

1 

1 

1 

1 

0 

a2 

62 

1 

a2 

0 

c2 

1 

b2 

c2 

0 

9. 


10. 


6 

15 

11 

10 

5 

16 

12 

9 

7 

14 

10 

11 

8 

18 

9 

12 

2 

3 

5 

9 

3 

7 

8 

11 

6 

10 

4 

2 

8 

4 

5 

10 

15. 


16. 


7    10 

13 

3 

14    19 

27 

6 

24   33 

41 

10 

31    47 

64 

15 

5    - 

■3 

-2 

4 

1 

-6 

-1 

4 

3 

0 

6 

-4 

11. 


a  6  c 

6  -a  0 

c  0  —  a 

0  c  6 


17. 


Ill 

y 

n 

X 

X 

y 

n 

m 

X 

71 

y 

m 

lit 

n 

y 

X 

I 

X 

X 

X 

X 

12. 

X 
X 

y 

X 

X 

y 

X 
X 

X 

X 

X 

y 

18. 


1 

1 

1 

b 

c 

d 

Ir 

c2 

d2 

V' 

c3 

d* 

XXIV.    SUPPLEMENTARY   TOPICS 

CUBE   ROOT 

252.  Cube  Root  of  a  Polynomial.  By  the  binomial  formula 
(§  179),  (a+b)3=a3+3  a2b  +  3  ab2+b3.  Any  polynomial  which 
may  be  put  in  this  form  is  a  perfect  cube.  Its  cube  root  may 
be  found  by  inspection. 


Example.     Find  v8  r3  +  36  r2  +•  54  r  +  27. 

Solution:    1.  8r3+36r2+,54r+27  =  (2r)3+3(2r)2  •  3  +  3(2  r)  •  32+33. 

2.    .-.  \/8  r8  +  36  r2  +  54  r  +  27  =  2  r  +  3. 

Notice  that  "  a  "  is  2  r  and  "  b  "  is  3. 

Note.     If  b  is  negative,  the  form  is  a3  —  3  d2b  +  3  a&2  —  63. 

EXERCISE   128 

Find  by  inspection  the  cube  roots  of : 

1.  8ar3  +  12x2  +  6x+-l.  4.    8t6-60t4-125  + 150?. 

2.  l-12a  +  48a*-64a3.  &     <t_a*b      aV_W. 

3.  27m6+-l+-27m4  +  9m2.  '8        4        6       27* 

253.  The  cube  root,  exact  or  approximate,  of  a  polynomial 
may  be  found  by  a  division  process. 

The  perfect  cube  polynomial  a3  +■  3  a2b  +  3  ab2  +-  b3  may  be 
put  in  the  form  a3  +-  b  (3  a2  -f  3  a&  +-  62).  This  expression  sug- 
gests the 

Rule.  —  To  find  the  cube  root  of  a  polynomial : 

1.  Arrange  the  polynomial  according  to  the  powers  of  some 
letter  (§  4,  /). 

256 


SUPPLEMENTARY  TOPICS  257 

2.  Write  the  cube  root  of  the  first  term  as  the  first  term  of  the 
root.  Cube  the  first  term  of  the  root  and  subtract  it  from  the 
given  expression. 

3.  For  the  trial  divisor,  take  three  times  the  square  of  the  first 
term  of  the  root.  Divide  the  first  term  of  the  remainder  (step  2) 
by  the  trial  divisor.  Write  the  quotient  as  the  next  term  of  the 
root. 

4.  For  the  complete  divisor,  add  to  the  trial  divisor  three  times 
the  product  of  the  new  term  of  the  root  by  the  part  obtained  previ- 
ously, and  also  the  square  of  the  new  term  of  the  root. 

5.  Multiply  the  complete  divisor  by  the  new  term  of  the  root 
and  subtract  the  result  from  the  remainder  (step  2). 

6.  Continue  in  this  manner  until  the  cube  root  or  the  desired 
number  of  terms  has  been  obtained :  (a)  for  the  trial  divisor,  take 
three  times  the  square  of  the  part  of  the  root  already  found; 
(b)  divide  the  first  term  of  the  last  remainder  by  the  first  term 
of  the  trial  divisor  for  the  new  term  of  the  root ;  (c)  form  the 
complete  divisor  as  in  step  4;  (d)  multiply  and  subtract  as  in 
step  5. 

«  Example  1.     Find  -i/8  x6  -  36  tfy  +  54  x*y2  -  27  tf. 

2x2-3y 


Solution:  1.   a  =  VS  oft  =  2  x2. 

2.  a3  =  8  x6  ;  subtract. 

3.  Trial  divisor  :  3  a2  =  12  x*. 
b  =-  36  x*y  +  12  x4  =  -  3  y. 
Complete  divisor :  3  a'2  =  12  x* 

3  ab  =  -  18  x2y 
b2  =  9  y2 


8  x6  -  36  x*y  +  54  x2y2  -  27  y* 
8x6 


3 a2  +-3  ab  +  &2  =  12x*-  18 tfy  +  9y* 
4.   Multiply  by  —  3  y.     Subtract. 


-36x42/  +  54xV-27y3 


36  x*y  -f  54  x2y2  -  27  y* 


258 


ALGEBRA 


Example  2.     Find  V28  0^-54 x+tf+3  x4- 9 z2-27-6  x*. 


s2-2x-3 


Solution:   1.    a—y/x^=x'\ 

2.  a3  =  x6 ;  subtract. 

3.  Trial  divisor:  3a2=3x4.     -QxB+3x*=-2x 
Complete  divisor :  3  a2 = 3  x4 

3a&  =  -6x3 
62=4x2 


x6-  6  x5 + 3  x4 + 28  x8-  9  x2-  54  x- 
x6 


-6x6+3x4  +  28x3 


3a2  +  3a&  +  62  =  3x*-6x3+4x2 

4.  Multiply  by -2  x.     Subtract. 

5.  Trial  divisor:  3a2=3(x2-2x)2  =  3x4-12x3+12x2 
6  =  -9x4-4-3x4=-3. 

3a&=3(x2-2x)(-3)  =  -9x2+18x 

52=(-3)2 +  9 

Complete  divisor:  3x4-12x3+3x2+18x+9 

6.  Multiply  by  -3.     Subtract. 


-6x5+12x4-8x3 


9x4+36x3-9x2-54x 


-9x4+36x3-9x2-54x 


EXERCISE   129 

Find  the  cube  roots  of : 

1.  ci  +  3c2d  +  3cd2+d\ 

2.  ri-3r2s  +  3rsi-si. 

3.  a6  +  12a4&  +  48a262  +  64&3. 

4.  27  ms +  135  m2n  + 225  inn2 +  125  n\ 

5.  a6-6ar5  +  9a;44-4ar}-9arJ-6a;-l. 

6.  8a?  +  36a5  +  66a4  +  63as  +  33a2  +  9a  +  l. 

7.  302/2+27^  +  12y-45?/4-8-35^  +  27  2/!. 

8.  9a3-36a  +  a6  +  21a4-9a5-8-42a2. 


254.  Cube  Root  of  an  Arithmetical  Number.  The  cube  root  of 
1000  is  10;  of  1,000,000  is  100;  etc.  Hence  the  cube  root  of 
a  number  between  1  and  1000  is  between  1  and  10 ;  the  cube 
root  of  a  number  between  1000  and  1,000,000  is  between  10 
and  100 ;  etc. 


SUPPLEMENTARY  TOPICS  259 

That  is,  the  integral  part  of  the  cube  root  of  a  number  of 
one,  two,  or  three  figures  contains  one  figure ;  of  a  number  of 
four,  five,  or  six  figures,  contains  two  figures ;  and  so  on. 

Hence  if  the  given  number  is  divided  into  periods  (§  62)  of 
three  figures  each,  beginning  with  the  units'  figure,  for  each 
period  in  the  number  there  will  be  one  figure  in  the  cube  root. 

255.  The  first  figure  of  the  cube  root  of  a  number  is  found 
b}r  inspection;  the  remaining  figures  are  found  in  the  same 
manner  as  the  cube  root  of  a  polynomial. 

Example  1.     Find  the  cube  root  of  157464. 

Solution  :  1.    157464  has  two  periods :   157  464.     There  are  In  the 
,  cube  root  two  figures,  a  tens'  and  a  units'  figure.  50  +  4 


2.  125000  is  the  largest  cube  in  157000. 
a  =  >y  125000  =  50.    Place  50  in  the  root. 
Subtract. 

3.  Trial  divisor  :  3  a2  =3(50)2  =  7500 
b  =  324  +  75  =  4+.    Place  4  in  the  root. 

4.  Complete  divisor :  3  ab  =  3  •  50  .  4  =   600 

62  =  42  =  _16 

5.  Multiply  by  4.  3  a2  +  3  ab  +  62  =  8116 


157  464 
125  000 


32  464 


32  464 


Rule.  —  To  find  the  cube  root  of  an  arithmetical  number : 

1.  Separate  the  number  into  periods  (§  62)  of  three  figures  each. 

2.  Find  the  greatest  cube  number  in  the  left  hand  period ;  write 
its  cube  root  as  the  first  figure  of  the  root ;  subtract  the  cube  of  the 
first  root  figure  from  the  left  hand  period,  and  to  the  result  annex 
the  next  period. 

3.  Form  the  trial  divisor  by  taking  three  times  the  square  of 
the  part  of  the  root  already  found  and  annexing  two  zeros. 

4.  Divide  the  remainder  (step  2)  by  the  trial  divisor  and  annex 
the  integral  part  of  the  quotient  to  the  root  already  found. 

5.  Form  the  complete  divisor  by  adding  to  the  trial  divisor  three 
times  the  product  of  the  new  root  figure  by  the  part  of  the  root 


260  ALGEBRA 

already  found,  with  one  zero  annexed,  and  also  the  square  of  the 
new  root  figure. 

6.  Multiply  the  complete  divisor  by  the  new  root  figure  and 
subtract  the  product  from  the  remainder. 

7.  Continue  in  this  manner  until  the  cube  root  or  the  desired 
number  of  decimal  places  for  the  root  has  been  obtained. 

Note  1.    Note  1,  p.  68,  applies  with  equal  force  to  the  above  rule. 
Note  2.     If  any  root  figure  is  zero,  annex  two  zeros  to  the  trial  divisor 
and  annex  the  next  period  to  the  remainder. 

Example  2.     Find  the  cube  root  of  8144.865728. 
The  solution  may  be  arranged  as  follows  : 

20.12 


8  144.865  728 
8 

120000 

600 

1 

120601 

121203 

120 

144  865 
120  601 

90 

;o 

4 

24  264  728 

121323 

54 

24  264  728 

Since  1200  is  not  contained  in  144,  the  second  root  figure  is  zero  ;  we 
then  annex  two  zeros  to  the  trial  divisor  1200,  and  annex  to  the  remainder 
the  next  period. 

EXERCISE  130 
Find  the  cube  roots  of  the  following  numbers : 

1.  19683.  t  4.  2515456.  7.   187149.248. 

2.  148877.  5.   857.375.  8.   444.194947. 

3.  59.319.  6.   46.268279.  9.    788889.024. 


SUPPLEMENTARY  TOPICS  261 

DETACHED  COEFFICIENTS 

256.  Detached  Coefficients.  Solutions  of  examples  in  "  long  " 
multiplication  and  division  may  be  abbreviated  as  in  the  fol- 
lowing examples. 

Example  1.     Multiply  3a?  +  2x-4:by  3  x-2. 


Solution:  (a) 

Solution  :  (6) 

3x3  +  0-x2  +  2a;     —    4 

3x3  +  0-x2  +  2x-   4 

Sx  -2 

Sx  -2 

9  z*  +  0  •  xs  +  6  x2    -  12  x 

9      +0         +6-12 

-6x3    -0-x'2-    4x 

+  8 

_6         -0-4     +8 

9  x4  -  6  x3    +  6  x2    -  16  z  +  8  9     -6        +6     -16     +8 

.\  Kesult = 9  z4- 6  a3 + 6  s2 - 16  a; + 8. 

Note  that  in  solution  (6)  only  the  coefficients  are  written 
in  the   partial   and  total   products ;  that   the  multiplier   and 
multiplicand  are  arranged  in  the  same  order  of  powers  of  x\ 
that  0  is  supplied  for  the  missing  powers. 
Solution  (6)  is  by  "  detached  coefficients." 
Example  2.     Divide  12  a3  -  25  a  -  3  by  2  a  -  3. 


Solution  :  (a) 

Solution  :  (6) 

6  a?  +  9  a     +1 

-3 

2a- 

6+9+1 

2  a  -  3 1 12  a3  +  0  •  a.2  -  25  a  - 

-3|12a3+   0-25a-3 

12  a3  -  18  a2 

12      -  18 

18  a2  -  25  a 

18-25 

18  a2  -  27  a 

18-27 

2a- 

-3 

2     -3 

2a- 

-3 

2     -3 

.*.  Result  =  6  a2  +  9  a  +  1. 
Solution  (b)  is  by  "  detached  coefficients." 

EXERCISE  131 

Solve  by  detached  coefficients : 

1-5.    Examples  21-25  on  page  12. 

6-10.   Examples  16-20  on  page  13. 

Note.    The  same  device  may  be  used  to  abbreviate  addition  and  subtrac- 
tion exercises. 


262  ALGEBRA 


PROOFS  OF  THE  RULES  FOR  THE  DIVISIBILITY  OF  a»  ±  6» 

257.  In  §  91,  the  rules  for  the  divisibility  of  an  ±  bn  were 
determined  by  inspection.  These  rules  may  be  proved  by 
means  of  the  factor  theorem. 

Proof  of  I,  1.  If  b  be  substituted  for  a  in  an  —  bn,  the 
result  is  bn  —  bn,  or  0.  Then,  by  §  94,  an  —  bn  has  a  —  b  as  a 
factor. 

Proof  of  I,  2.  If  —  b  be  substituted  for  a  in  a*  —  bn,  the 
result  is  (—  b)n  —  bn.  When  n  is  even,  (— b)n  —  bn  =  bn  —  bn  =  0. 
Then,  by  §  94,  an  —  &n  has  a  —  (  —  b)  or  a  +  &  as  a  factor,  w;^ew 

Proof  of  I,  3.  If  b  be  substituted  for  a  in  an  ■+-  bn,  the 
result  is  6n  +  6n,  or  2  6".  This  result  is  not  zero  unless  tr  is 
zero.     Then,  by  §  94,  an  +  bn  never  has  a  —  b  as  a  factor. 

Proof  of  I,  4.  If  —  b  be  substituted  for  a  in  on  -f  6n,  the 
result  is  (-  b)n  +  &n.  When  n  is  odd,  (— 6)n^6M=  —  bn  +  6n=0. 
Then,  by  §  9.4,  an  +  6n  has  a  —  (  —  b)  or  a  +  6  as  a  factor,  wAew 
n  is  odd. 

258.  The  Highest  Common  Factor  of  Polynomials  which  can- 
not be  Readily  Factored.  The  rule  in  arithmetic  for  finding 
the  H.  C.  F.  of  two  numbers  is : 

1.  Divide  the  greater  number  by  the  less. 

2.  If  there  is  a  remainder,  divide  the  divisor  by  it.  Continue 
thus  to  make  the  remainder  the  divisor  and  the  preceding  divisor 
the  dividend,  until  there  is  no  remainder. 

3.  The  last  divisor  is  the  H.  C.  F.  required. 


SUPPLEMENTARY  TOPICS  263 

Example.     Find  the  H.  C.  F.  of  169  and  546. 


169)546(3 
507 
39)169(4 
156 


the  H.  C.  F.  of  169  and  546  is  13. 


13)39(3 
39 

A  similar  process  serves  for  polynomials. 
Let  A  and  B  be  two  polynomials,  the  degree  (§  18)  of  A 
being  equal  to  or  greater  than  that  of  B. 


Suppose  that  B  is  contained  in  A  p  times, 
with  a  remainder  (7;  that  C  is  contained  in 
B  q  times,  with  a  remainder  D ;  and  that  D 
is  contained  in  G  exactly  r  times. 


B)  A(p 
pB 

C)  B(q 
qC 

D)  C(r 
rD 
0 


Then  D  is  a  common  factor  of  A  and  B. 
Proof.     Since  dividend  =  divisor  x  quotient  +  remainder  : 

A=pB+C.        (1)  B=qC+D.        (2)  C  =  rD. 

Substitute  the  value  of  C  in  (2)  ;  then, 

B  =  qrD  +D  =  D(qr  +  1).  (3) 

Substitute  the  values  of  B  and  C  in  (1)  ;  then, 

A  =  pD(qr  +  1)  +  rD  =  D(pqr  +p  +  r).  (4) 

From  (3)  and  (4),  D  is  a  common  factor  of  A  and  B. 
Further,  every  common  factor  of  A  and  B  is  a  factor  of  D. 
Proof.     Let  F  be  any  common  factor  of  A  and  B  ;  and  let 

A  =  mF  and  B  =  nF. 
Then  :  from  (1)         C  =  A  -  pB  =  mF  -  pnF,  (5) 

from  (2)         D  =  B-qO.  (6) 


264  ALGEBRA 

Substituting  in  (6)  the  values  of  B  and  C, 

D  =  nF  -  q(mF- pnF)  =  F(n  — qm  +  qpn).  (7) 

Hence  F  is  a  factor  of  D. 

Then,  since  every  common  factor  of  A  and  B  is  a  factor  of 
D,  and  since  D  itself  is  a  common  factor  of  A  and  B,  it  follows 
that  2)  is  the  highest  common  factor  of  A  and  B. 

In  applying  the  process  to  polynomials  the  following  notes 
should  be  observed. 

Note  1.  Each  division  should  be  continued  until  the  remainder  is  of  a 
lower  degree  than  that  of  the  divisor. 

Note  2.  If  the  terms  of  one  expression  have  a  common  factor  which  is 
not  a  common  factor  of  the  terms  of  the  other  expression,  the  factor  may  be 
removed,  for  it  evidently  cannot  form  part  of  the  common  factor  of  the  two 
expressions.  In  like  manner,  any  remainder  may  be  divided  by  a  factor 
which  is  not  a  factor  of  the  preceding  divisor. 

Note  3.  If  the  given  expressions  have  a  common  factor  which  may  be 
seen  by  inspection,  remove  it  and  find  the  H.  C.  F.  of  the  resulting  expres- 
sions. The  result  multiplied  by  the  common  factor  that  has  been  removed  is 
the  H.  C.  F.  of  the  given  expressions. 

Note  4.  If  the  first  term  of  the  dividend,  or  of  any  remainder,  is  not 
divisible  by  the  first  term  of  the  divisor,  it  may  be  made  so  by  multiplying 
the  dividend  by  any  number  which  is  not  a  factor  of  the  divisor. 

Example  1.     Find  the  H.  C.  F.  of 

6a?-25x2  +  Ux  and   6 ax2  + 11  ax- 10 a. 

Solution  :  1.  Kemove  X  from  the  first  expression  and  a  from  the 
second.     (See  Note  2.)     Then  continue  as  below. 

6  s2  -  25  x  +  14 1 6  s2  +  1 1  x  -  10 1_1 
6  s2-  25s  +  14 
Divide  by  12.     (Note  2.)  12|36s-24 

3x-    2[6s2-25s  +  1412s-7 
6  s2-    4x 

-21s+  14 
.-.  3s-2istheH.C.F.  -  21  x  +  14 


SUPPLEMENTARY  TOPICS  265 

Example  2.     Find  the  H.  C.  F.  of  2  m3  -  3  m2  -  8  m  -  3, 
and  3  m4  —  7  m3  —  5  ra2  —  m  —  6. 

Solution  :  Since  3  m4  does  not  contain  2  m3,  multiply  the  second  ex- 
pression by  2.     (See  Note  4.) 

8  m4  —  7  m8  —  5  m2  -  m  —  6 

2 


2m8-3m2-8m-3|6m4-14m3  —  10m2 -2m-  12  [3m 
6m4-    9m3 -24m2 -9m 

-  5m3  +  14m2  +  7m -12 
--2 

10  m8  -  28  m2  -  14  m  +  24  |£ 
10m8 -15m2 -40m -15 

-  13|  -  13m2  +  26m  +  39 

m2  -  2  m  -  3 
w2_2m-312m3-3m2-8m-3  [2ro-l 
2  m3  -  4  m2  -  6  m 

m2-2m-8 
m?  —  2  m  —  3 


.-.  m2-  2m  -  3  is  the  H.C.  F. 

Notice  that  —5  m8  of  the  first  remainder  does  not  contain  2  m8,  and 
that  the  remainder  is  therefore  multiplied  by  —  2.  Notice  also  that  the 
divisor  —  13  is  removed  from  the  second  remainder,  thus  making  the  first 
term  of  the  new  divisor  positive. 


EXERCISE  132 
Find  the  H.C.F.  of: 

1.  ar>+5a>-24  and  ar*  +  4  a2  -  26  a> -f  15. 

2.  3x2-±x-4:im&3xi-7xB  +  6xl-9x  +  2. 

3.  2  m4  +  5  m3  —  2  m2  -f  3  ra  and  6  ra3w  —  7  m2n  4-  5  mn  —  2  ^. 

4.  x*y—6xy  —  27y  and  x^y  —  2x?y  —  8xy  +  21  #. 

5 .  4  £y  - 15  xy2  +  9  ?/3  an  d  8  xA  -  18  arfy  +  25  afy2  - 1 2  a^/8. 

6.  3  n8  +  8 w2 - 9 n  +  2  and  6rc4  +  23 w3  +  2 w2-  13 n  +  2. 


266  ALGEBRA 

7.  6a6+5a5-6a4-3a34-2a2and9a4  +  18a3+5a2-8a-4. 

8.  3&4-13&3  +  3&2  +  4&and9&3  +  12&2-86-5. 

9.  12  a3-5  a2x~ll  ax*+6x3  and  15  a3+ 11  a2x-  8 aa2-  4  x3. 
10.  2ar}-3ar  +  2x-8  and  3^-7^  +  4^-4. 

259.  The  L.  C.  M.  of  Two  Polynomials  which  cannot  be  readily- 
Factored.  Let  A  and  B  be  two  polynomials ;  let  F  be  their 
H.  C.  F.  and  M  their  L.  C.  M.     Let  A  =  aF  and  B  =  bF. 

Since  F  is  the  highest  common  factor  of  aF  and  bF,  a  and 
6  cannot  have  any  common  factors.  Hence,  the  L.C.M.  of  aF 
and  bF  is  abF. 

That  is,  M=abF=a(bF)  =  aB; 

or  j*f  =  abF=  b  (aF)  =  bA. 

Rule.  —  To  find  the  L.  C.  M.  of  two  polynomials : 
Divide  one  of  the  polynomials  by  their  H.  C.F.  and   multiply 
the  quotient  by  the  other  polynomial. 

EXERCISE  133 

Find  the  L.C.M.  of: 

1.  3a2-13a  +  4and3a2  +  14a-5. 

2.  6a2  +  25«Z>  +  2462and  12a2  + 16a& -362. 

3.  12  m2  —  21  m  —  45  and  4  m3  — 11  m2  —  6  m  -f-  9. 

4.  2a3-5a2-18a-9and3a3-14a2-a  +  6. 

5.  6a3-7a2  +  5x-2and4»4-5a2  +  4a-3. 


INDETERMINATE  FORMS 

—   becomes  -   for  .. 
x-3  0  x-3 


260.  The  fraction   -  becomes  -•   for  #  =  3;  - 

comes   -•     Neither  has  any  meaning,  for  division  by  zero  is 

not  allowed  (§  1,  a).  Results  like  these,  however,  must  be 
interpreted  at  times.  The  following  paragraphs  show  how  to 
give  the  interpretation. 

261.  A  constant  is  a  number  which  always  has  the  same 
value  in  a  particular  mathematical  discussion. 


SUPPLEMENTARY   TOPICS  267 

A  variable  is  a  number  which  assumes  different  values  in  a 
particular  mathematical  discussion.  * 

Thus  n  may  assume  the  values  .1,  .01,  .001,  •••,  etc. 

A  limit  of  a  variable  is  a  constant  the  difference  between 
which  and  the  variable  may  be  made  to  become  and  remain 
less  than  any  assigned  positive  number,  however  small. 

Thus,  the  variable  n  above  is  evidently  approaching  the  value  0 ;  or, 
the  limit  of  n  is  zero. 

The  symbol  =  is  read  "  approaches  the  limit."  Thus,  n  ==  0  means 
"  n  approaches  the  limit  zero." 

262.  If  a  number  becomes  and  remains  greater  than  any 
positive  number  which  may  be  assigned,  it  is  said  to  become 
infinitely  large  or  to  approach  infinity  as  limit. 

The  symbol  oo  is  called  "  infinity." 

Thus,  if  n  represents  any  positive  integer  (assuming  therefore  the 
values  1,  2,  3,  •••,  etc.),  it  approaches  infinity  as  limit ;  i.e.  limit  of  n  =  oo, 
or  n  =  oo. 

Note,  co  is  not  a  symbol  for  some  definite  value.  It  is  a  symbol  for  the 
limit  of  a  number  which  "  becomes  and  remains  larger  than  any  assigned 
positive  number." 

Evidently  as  n  =  oo,  also  n2  =  oo.  ^^t  n2  =  cc  is  read  "  the 
limit  of  n2  as  n,  approaches  oo  is  infinity." 

263.  Interpretation  of  -.     To  determine  the  meaning  of  -, 

replace  -  by  -  and  consider  limit  -  as  x  =  0. 
K)       x  x 

If  x  becomes  .1,  .01,  .001,  ...,  etc.,  1  becomes  10,  100,  1000,  •••,  etc. 

x 

Evidently,  then,  -  increases  indefinitely.     That  is,  hmit  I  =  ».     Then, 
x  *=M)  x 

to  the  otherwise  meaningless  form  - ,  give  the  value  oo. 


270  ALGEBRA 

EXERCISE   134 

Find  the  values  of  the  following  as  x  =  0*. 

1.    t       2-    £•       3     J_-       4.    7;»«      .       5.  * 

a;  a2  /1\  a;  (a +  5)  #(#  +  1) 


Tind  the  values  of  the  following  as  #  =  oo  : 

6.    x2.  7.    2*.  8.    ?•  9.    2  +  -.  10.    i« 

a;  a;  2X 


Find  the  values  of  the  following 
mit/ 

-2  v^2 


11.    limit/'      ^~4      V  14     1imit/^-^-6Y 


12.  limit (fit&\.  15.    limit/1 1.,  ,g V  • 

»-°°  \    2/     /  '^  Vn      n(n  —  5)J 

13.  limit  fld^Y  16     li^tA24-2tt-7\ 

17.  The  equations  ?/  =  2  sc  -f  3  and  y  =  2  x  -+-  5  have  no  com- 
mon solution  according  to  §  50.  Consider  y  =  2  a;  -\-  3  and 
?/  =  aaj-}-5.  Solve  them  as  simultaneous  equations,  and  hnd 
the  values  of  x  and  y  as  a  =  2. 

18.  Solve  2^  +  3?y  =  6  and  £x  +  by  =  7  as  simultaneous 
equations,  and  find  the  values  of  x  and  y  as  6  =  6. 

§  267.  Graphical  Solution  of  Equations.  In  §  95,  the  state- 
ment was  made  that  an  equation  of  the  nth  degree,  having  one 
unknown,  has  n  roots,  but  that  these  roots  are  not  readily  found 
for  equations  of  degree  above  the  second.  Such  equations  may 
be  solved  graphically  as  in  §  75. 


SUPPLEMENTARY    TOPICS 


271 


Example.     Solve  the  equation  x?  —  4  #2  —  2  x  +  8  =  0. 


Solution  :     1.     Let  y  =  x3  —  4  x2  —  2  x  4-  8. 

When  x  =      0      1           2          34        5-1        -2        -8 

then  y  =      83      -4      -70     23          5      -12      -49 

X7~ 

*^     s 

^          \ 

y            X 

v     -IE         V 

7-               5                ^ 

-/                 \                          -,r 

/--      I. L                                3 

a-C                 ^    _         C7                               £ 

A/~                        0                ^JB                                       -fcf- 

r        zl             -?      7     -                                   ^      ?                         "fr      "!x 

—                    —         t                                                   \                                     t 

V                         4 

1                                                                                                       V                                            £ 

/                              75                                            S^                       / 

it                               t                                                               %            ^ 

s^ta5* 

X                        dfc 

f.                          -iy. 

J 

/                       T7 

t             is.r 

2.  The  curve  crosses  the  horizontal  axis  at  points  A,  B,  and  C.  Hence 
its  roots  are,  approximately,  —  1.42,  +  1.42,  and  +  4. 

Check  :  This  equation  may  be  solved  as  in  §  95.  Using  the  factor 
theorem  :    Xs  -  4x2  -  2x  +  8  =  (x  -  4)  (x2  -  2)  =  0. 

.-.  x  =  4;  also  x2  =  2,  or  x  =  ±  V2  =  ±  1.414. 

Clearly  the  results  ±  1.42  obtained  graphically  are  close  to  the  roots 
±  1.414. 

EXERCISE  135 

Solve  the  following  equations  graphically : 

1.  a;3-3a;2-#4-3  =  0.  4.   a;4  - 10  x*  -f  9  =  0. 

2.  or5 -4  a2 -7  a +  10  =  0.  5.    x?  +  x2-  10  £-  10  =  0. 

3.  ot'  +  jb2  — 6a  =  0.  6.   a?  — ^-8  a  +  8  =  0. 


272  ALGEBRA 

268.  Horner's  Synthetic  Division.  Synthetic  division  in  the 
case  when  the  divisor  is  a  binomial  is  considered  in  §  93.  A 
similar  process  of  division  may  be  employed  in  other  cases  of 
division  of  a  polynomial  by  a  polynomial. 

Consider  :  3  x2  —  2  x  +  4 

2x2+x 


6x* 
6x* 

-       X3- 

+  3x3- 

-3x2 
-9x2 

+ 

10  x- 

12 

-  4  x3  +  6  x2 

-  4  x3  -  2  x2 

+  10  x 

+    6x 

8x2 
8x2 

+ 
+ 

4x- 
4x- 

12 

12 

6x4 

-    x3  -  S  x2 

+  10  x 

-  12 

-  3  x3  +  9  x2 

-    6x 

+  2x2 

-    4x 

+  12 

When  performing  the  subtractions,  the  signs  of  the  terms 
subtracted  are  changed  and  the  results  are  added  to  the  minu- 
ends. If  +  x  and  —  3  of  the  divisor  are  changed  in  advance 
to  —  x  and  4-  3,  the  various  partial  products  may  be  added  to 
the  respective  remainders.  Following  this  suggestion  and  pro- 
ceeding in  a  manner  entirely  similar  to  that  in  §  93,  the  solu- 
tion may  be  arranged  as  follows : 

2x2 
—  x 
+  3 

3x2-2x  +  4     ||      0+0 

Steps  of  the  process. 

1.  Change  the  signs  of  all  terms  of  the  divisor  excepting  the  first  term. 

2.  Divide  6X4  by  2  x2,  getting  3x2,  the  first  term  of  the  quotient. 
Multiply  —  x  +  3  by  3  x2,  getting  —  3  x3  +  9  x2,  which  are  written  in  their 
proper  columns. 

3.  Add  the  second  column,  and  divide  the  result  by  2  x2,  getting  —  2  x, 
the  second  term  of  the  quotient.  Place  this  second  term  of  the  quotient 
at  the  foot  of  the  second  column  below  the  line. 

4.  Multiply  —  x  +  3  by  —  2  x,  getting  2  x2  —  6  x,  which  are  written  in 
their  proper  columns. 

5.  Add  the  third  column  and  divide  the  result  by  2  x2,  getting  4,  the 
third  term  of  the  quotient.  Place  this  third  term  at  the  foot  of  the 
third  column  below  the  line. 


12  x* 

-  11 

x2y  + 

Oxy* 

-9y3 

+  8 

- 

16 

- 

2 

+  4 

SUPPLEMENTARY    TOPICS  273 

6.  Multiply  —  x  +  3  by  4,  getting  —  4  £  +  12,  which  are  written  in 
their  proper  columns.  Add  the  remaining  columns  and  thus  find  the  re- 
mainder.    In  this  example,  the  remainder  is  zero. 

The  quotient  is  3  x2  —  2  x  -f  4.  This  is  observed  to  agree  with  that 
found  previously. 

Example  2.     Divide  12  x*  -  11  x2y  -  9  f  by  3  x2  -  2  xy+4  y2. 

Solution  : 

3x* 

+  2xy 

4  x  -     1  y  ||  -  18        -  5 
Quotient :  Ax—  y.     Remainder  :    -  \&xy'z  —  5j/3. 

EXERCISE   136 

Divide  the  following  by  synthetic  division  : 

1.  12z3-7.T2-23a-3by  4.t2-5z-3. 

2.  4  a4  -  9  a2  +  30  a  -  25  by  3  a2  +  2  a  -  5. 

3.  2  a4  -  a3  +  8  a  -  25  by  2  a2  -  3  a  +  5. 

4.  4  m2  +  1  4- 16  m4  by  2  m  +  4  m2  +  1. 

5.  6  a5  -  13  xA  -  20  or5  +  55  a2  -  14  x  -  19  by  2  a2  -  7  a  +  6. 

6.  8a5-4a42/-8xy-18a2/4  +  2l2/5  by  ±x? -2x2y +  6xy2 
-If.       , 

7.  37a2  +  50  +  a5-70aby  2a2  +  5  +  a3-6a. 

MISCELLANEOUS  EXAMPLES 
A.    The  Four  Fundamental  Operations. 

1.  Add  3(a  -  6)2  -  9,  4(a  -  b)2  -  5 (a  -  b),  and  -7(a-6)2 
+  8(a -6). 

2.  Simplify  6  ran  +  5  —  ([  —  7  m*  —  3]  —  \  —  5  mn  — 11  j ). 

3.  Simplify  7x-(5x-[-  12a;  +  6  as  -  11]). 

4.  Subtract  (2  a  -  3  %2  from  (5  a  -  4  %2. 

5.  Subtract  (jp  -+-  (?)#  from  mx. 


274  ALGEBRA 

6.  Multiply  a3pbq,  b4cm,  and  cna2p. 

7.  Multiply  7  xmy4n  —  8  xhyp  by  —  3  x*yn. 

8.  Multiply  x2p+*y  —  x7yq  by  a;2"-1  +  yq~x. 

9.  Multiply  4  aw+562  -  3  a4bn  by  aw+26  -  2  aft""1. 

10.  Multiply  am  +  bn~  cp  by  aw  —  bn  +  cp. 

11.  Simplify  (a  +  2  fc)2  -  2  (a  +  2  6)(2  a  +  6)  +  (2o  +  b)\ 

12.  Simplify  (a;  +  y  +  z)3  -  3(#  -J-  z)(z  +  a-)(a;  +  y)- 

13.  Divide  am+16n+3  by  —  ab2. 

14.  Divide  xp+qyn+2  —  a?+y  by  a5?/3- 

15.  Divide  a7pbqcir  —  ahpbZqc2r  —  a3p65«c3r  by  —  aZpbqc2r. 

16.  Divide  a;3w+2  +  8  a3™-1  by  x2m+l  -2x2m  +  ±  x2m~K 

17.  Divide  xim  +  a?2m?/4n  -f  ySn  by  #2m  -  scmy2B  +  2/4n- 

18.  Divide  x3  +  (a  —  6  —  c)x2  +  (—  ab +  bc  —  ac)x  +  abc  bjKx2 
-f  (a  —  b)x  —  ab. 

19.  Divide 

a(a  —  b)x2  +  (—  ab  +  62  +  6c)a:  -  c(6  +  c)  by  (a  -  &)#  -f  c 

20.  Divide    x?  -  (3  a  +  2  &  -  4  0)^+ (6  a&  -  8  6c  + 12  ca)a  - 
24  abc  by  x  —  2  b. 

Set  B.     Fractions. 
Simplify  the  following : 

s«  +  2g'-8a;-16  4    (2  x2  +  5  a?-2)2-25 

"  ^^a^  +  Saj-lG*  .'   (3  a;2 -4  a;- 3)2- 16* 

"2^  +  37      ¥i~27.  5.   ^"J^tl^i"1'        • 

2'   4^-92/2+9a:2-42/2  "+"  + 


Q    a;2  —  5x  —  84  #    a?  -f-  7  fl  y  —  z  z  —  x 

o. 


27  a;3 -8         3  a; -2  tf-ty-z)*      (x-z)2-: 


SUPPLEMENTARY   TOPICS  275 


3  3  5n2  5n2 

'  2»  +  l     2»--l      8n3  +  l      8?i3- 

a  1  3      _  a2  +  2  a 

a  +  3     a-3~a2-9~~  a2  +  9  ' 

_      3  a         3  a         6  a2         12  a4 

y-   — 7~r  H r  +   o  .  ro  +  "3 


a  +  6     a  — ft     a2  +  62     a4  +  64 

10  3  1  a- 2      2  a3 +  4 

2(a-l)      2(a  +  l)      a2  +  l       a4-l 

11.— 1 t^+  1 


2  a2  +  3  a  -  2      3  a2  +  o  a  -  2      1  +  a  -  6  a2 
_2 1         1 2__ 


?/ 2 # 2 

2a  +  ?/  »  +  2y 

__      a^-2a;2-4a;4-8     .  ^c2  +  9  a?  +  14  „  s2  -4  3?  +  ^ 

lo.    — i 7z — ;; tcz Tcz — ~  I  — i ~ tc~  X 


x4  +  3ar}-27a;-81      Vaj2  +  6i«  +  9  ^2 

■'         6  a2  -  a  -  2      ^  8  a2  -  18  a  -  5  v,       4  a2- 

■1-4.      ■■       t         t~t  ~  /n  t~t      ~  ~  ~  X 


4a  +  4\ 

-9  ; 


15. 


4  a2  -  16  a  +  15      12a2-5a-2     4  a2  +  8  a  +  3 

fx±l\2  _  fx-l\2 

Set  C.     Linear  Equations.     (One   Unknown.) 

1.   Solve  the  equation +  — — -  =  2. 

Divide  each  numerator  by  its  corresponding  denominator  ;  then 

1  +  -^—  +  1  -  ^^  =  2 ,  or  _?_  -  »±±  =  0 . 
2x-3  z2  +  4  2x-3      &  +  4 

Complete  the  solution. 


274  ALGEBRA 

6.  Multiply  a3p6«,  b4cm,  and  cn«2p. 

7.  Multiply  7  xmy**  —  8  ^p  by  —  3  afyB. 

8.  Multiply  a^+fy  —  a?y  by  a^"1  +  if~x. 

9.  Multiply  4  aw+562  —  3  a46*  by  cr+26  -  2  aM"1. 

10.  Multiply  am  +  bn  —  cp  by  aw  —  bn  +  cp. 

11.  Simplify  (a  +  2  6)2  -  2  (a  +  2  6)(2  a  +  6)  +  (2a  +  6)2. 

12.  Simplify  (a;  +  #  +  zf  -  S(y  +  z)(z  +  x)(x  +  2/). 

13.  Divide  am+1bn+*  by  —  db2. 

14.  Divide  x*+qif+2  —  x*+hf  by  a*?/. 

15.  Divide  dIpb*c4r  —  a^b^c27  —  a*?b5*<?r  by  —  a*"b''c2r. 

16.  Divide  x****  +  8  a^"1  by  x2m+l  -2x2m  +  4  x2"-1. 

17.  Divide  x^  +  a^y4*  +  3/**  by  af?  —  #m2/2rt  +  2/4n- 

18.  Divide  a?  +  (a  —  6  —  c)ar2  +  (—  ab  +  be  —  ac)  x  +  abc  by  x2 
+  (a  —  £»)a;  —  a& 

19.  Divide 

a(a  —  b)x* +  (—  a&  +  ff  -f  ^c)ar  —  c(6  +  c)  by  (a  -  b)x  +  c. 

20.  Divide    a?-(3a  +  2b-4c)x2+(6ab-8bc  +  12  ca)x- 
24  ate  by  a;- 2/,. 

#e£  J5.     Fractions. 
Simplify  the  following : 

x*  +  2a?-$<t-16  (</x'  +  r>x-2)2-25 

'  x*-2x*  +  Hx~16'  '   (Xx2~-4rX-'S)2-16' 

~7» — f~  o —      "o 71 —  ■    a?  —  2a^-f-2a; —  1 

2  a;     3  y        3  a?     2  y  .  5-  -j — ~— : 

Z'   4a?-<J;y2"f9a:2-4;,/!  *  +  *  + 

3    a?-5a?-84  ,    .r-}-7_  y-g z-x 

3x-2'  '  0-Qt-zY     (x-zf-tf' 


SUPPLEMENTARY   TOPICS  275 

3  3  5n*  5n2 

2w+l      2m-1      8n3  +  l      8ft3-l* 

a  1  3         a2  +  2« 


a  +  3     a-3     a2-9       a2  +  9 


9      3a         3  a         6  a2         12  a4 

a  +  b     a  -  b     a2+62     a4  +  &4 ' 

10    __3 1  a-2      2a'  +  4 

2(a-l)      2(a+l)      a2  +  l       a4-l 

1  1  1 


11. 


2z2  +  3a;-2      3^  +  5a;-2      l  +  a;_6a? 
_2 1         1 2_ 

12.  ^-t-//    g      .v    x  +  y 


XV                             XV 
y 2 x ^ 

2x+y  x+2y 


x*-2x*-<±x  +  8    ^_  /a2  +  9  a:  +  14     a?-4a?  +  4\ 
iC4  +  3^-27a;-81^"\ic2-f  6a;  +  9  >       x*-9     J 

6a»-g-2      vy8a2-18a-5  w       4a2-9 


4a2-16a  +  15      12a2-5a-2     4a2  +  8a  +  3 

fx+i\*   As-iy 


>Se£  C     Linear  Equations.     (One  Unknown.) 

1.   Solve  the  equation  - — ^—  +     ~  ,  =2. 
2»-3     a^H-4 

Divide  each  numerator  by  its  corresponding  denominator  ;  then 

l  +  -^_+l-^±i=2,or-l_-^±l  =  0. 
2x-S  z2  +  4  2x-3      a? +  4 

Complete  the  solution. 


276  ALGEBRA 

Solve  the  following  equations  : 
2. 

3. 


5. 


6. 


11. 


12. 


x  +  t     x+2 

_  x  +  5     x  +  6 

x+2     x+3 

x  +  6     a-  +  7 

8            3 

10           5 

x+3     x-7 

x  +  9     x-2 

x  +  ?>      x  +  4 
x+2     x+3 

,  x  +  2      Q 

3  +  2  = 

X  +  9           X  +  4: 

=     *     +      4      . 

a  +  3      a; +  18 

2«  +  3     2x- 

3          36      _0 

„      a  jo  -f  u       oart  ^*  •*'  ~r  -1  «? -i 


2aj-3 

2cc  +  3      4ar>-9 

2a;  +  5 

3ar>  +  24a;  +  19 

a; 

+  7 

x2  +  8  a;  +  7 

ar* 

a^ 

-2a? 
-2a> 

+  5     a£  +  3»-7_ 

-3      ar>  +  3a;  +  l~ 

5 

1              10 

g>     ~ ^-ri,    .   «,  t^ L  =  2. 


9. 


2a>-l      6z+5     3a;-4     4x  +  l 
q  +  6  ,  a  —  26_(2a—  6)a?  +  3a6 


10.    r_L^  + 


x  x  +  a  xr  —  a 


bx     a2  +  b2     a2 
a          a2          b2 

x(a  —  b) 
b 

a            b         a 

-b 

a     x 


13     a(x-a)  +  b(x-b)  =  a  +  b 
x  —  b  x  —  a 

14.    (a  +  b)(x-a  +  b)-(a-b)x  +  a2- b2=2a(x  +  a-b). 


SUPPLEMENTARY   TOPICS 

15.    (x  +  p  +  q)  (x  —  p  +  q)+  q2  =(x  —  p)  (x  +  q). 
117  3 


277 


16. 


17. 


x  —  2  a     6  a?  -+-  a     3  a;  —  8  a     2x  —  Sa 

4  1  _4 1 

« —  4  n      x  +  n      a;  +  4w      «  +  3w 


ar*  —  2  aa;  +  ct2       ar*  -f  ax  —  2  a2  _  o 
a~>  _  2  a  x  -  3  a2     ic2  +  az  +  2  a2 

ff-f-a  ,  a;  +  5  ,  x  —  a  —  &_q 

19.  (  -  -|         -         -    - — o. 

x—  a     x  —  o     x-{-a-\-b 

20.  ar3  +  (a;  -  a)3  +  (x  -  bf  =  3x(x-a)(x-  6). 


#e£  Z>.     Linear  Equations.     (Two  or  More   Unknowns.) 
Solve  the  following  pairs  of  equations : 

x  +  y 
4. 

X  -  1  y  +  4  =         ^ 


2. 


3. 


a?  +  H   ■  y-6=      d 
7  5 


a;  — 


2  10 

11 


a;  —  y  10 

3x  +  8     6a;-l 


9      a^+5_      o?/ 
2         ^~-~3^ 


2a;— 3y  4a;+6y__      1 

4  3     '  ~     2' 

5a  +  2y  7y-3a;  =  39 

2  5           10* 


5. 


2/_4       22/  +  ^ 

3 

5  (a;  -  4  y)     4(a?  -ff)=16 
6  9 


8a?-3     y-5^y 
4  3         6 


a;  — 


y- < 


|(2a;-l)(2/-4)-(a;-o)(2^4-5)=121. 
'1  4a*-3y=-29. 


278  ALGKBRA 

8      fK/>-<7)-Hl>-3<7)  =  9-3. 
1    f(f»-S?)+f(P  +  ?)=18. 

5  a;  —  4  y  _  _  3 
5x  +  42/_      13* 

5  *  —  s  +  .y-j. 

a?-4,y +6  1 

8aj-2y-18  4* 


10. 


Solve  the  following  for  x  and  7/ : 

f  a&(a  —  &)«  +  ab(a  +  b)y  =  a2  +  2  a&  —  62. 


12. 


13. 


14. 


15. 


16. 


'  ax  +  by  =  2. 

ra(a  +  y)  +  &(«  - 2/ )  =  2. 
7tt2(#  +  y)  —  n2(x  —  y)=m  —  n. 

(a  +  b)x+(a  -  b)y  =  2(a2  +  b2). 
b  a 

x  —  a  —  b     y - a  +  b 

(a  +  b)x  +  (a-b)y  =  2a2-2b\ 
y  x  4  ab 

a  —  b      a  +  b      a2  —  b2 


bx  +  ay  =  2. 
+  b)x  -  ab(a  -  b)y  =  a2  +  &2. 


{a6(a 
y  —  a  +  b  _y  —  a 


x  +  a+b      x  +  b  |  ay  —  bx  =  a2  +  b2. 

a-x     b  '     {(a  +  b)x  +  (a-b)y=2a2-2b2. 


y-b      a 

x-\-(a- 

b2)y  =  2a2  +  2b2. 


18      (      (a  + &>*  +  («-%=  2  a. 
l(a2-&2)a;+(a2 


SUPPLEMENTARY   TOPICS  279 

Set  E.     TJieory  of  Exponents. 
Square  the  following  by  the  rule  of  §  10,  b  : 
l.   3  a* +  4  6""*.  2.   5m-¥-8m2r4. 

3.  Square  a?b~*  -  2  at  —  a~lb%. 

4.  Expand  (4:X*y~% +  7  z~2)(4:X*y~* -7  z~2)  by  the  rule  of 
§  10,  a. 

Find  the  value  of : 

5.  25n^-49wl,  by  the  rule  of  §  15,  6. 

5  a-3- 7  m* 

6.  8g2+27^,  by  the  rule  of  §  15, /. 
2  x*  +  3  jf* 

a"-*-,  8_   al-H.     (See  §91:  1,2.) 

#*  —  x  5  a3  -t-  b  6 

9.  (3^-4^)3.  10.  (cr263  +  2a36-2)3. 

Find  the  square  roots  of  the  following : 

11.    16a-«mK  12.    49<bV"*.  13. 


4  6  V3 

14.  9  a**  -  6  a;*  +  25  -  8  af*  + 16  af*. 

15.  4  a-*  +  20  a-* -f  21  tT^- 10  cT^  +  1. 

16.  «  V3  -  6  aV2  +  5  6-1  +  12  of*  +  4  a~*&. 

Find  the  cube  roots  of  the  following : 

17.  9tff,m  18.    _64a-^M.         19.   *»$>. 

20.  27  »*  +  54  a^  +  36  a?V"*  +  8  2T2- 

21.  a?*  -  6  a;*  +  2.1  x~*  -  44  af  *  +  63  aH  -  54  af*  +  27  af  i 


280  ALGEBRA 

Simplify  the  following,  expressing  all  the  results  with  posi- 
tive exponents  : 


22.   [^(^2/-2)-^(^V)]T1.    23.    -_^X-X^_ 

V64Vc         Va8 

24.  [a""1  x  (a-1  n+1]  X  [(a~n)_1  X  (a"-2)-1]. 

25.  (a?  «    Xas^)«.  si.   fL±A_  _«!+&!• 

a*  +  &1        a  —  b 


m+n\  2m  /n2m\  m—n 

26.  r=— 1     ?-)     .  i      i      _i     ._i 


/am+n\2m/a2m\ 


n+1  n-1     n-1 


32.     Z-2IS-   + 


a*  —  $      a  2  —  6  2- 


27.  (an_1-j-an+1)  2n  .  **  &» 

2„  _  1  X2n  +  1 


1  1        mn  1 


33. 


p-q  p 


28.  (a;",-s-a;",),,l+H-*-a!   n-  *R  +  1       x2»-l 

34       <»*+&*     x°~*  +  6~*   11 

29.  [^(•M)]»    •  ^  a-l_^X    ai_6i    + 

ai  -i-  y\     x  +  y  a*+2  6*        7  a*6*  +  6  6* 

a-i  —  yi     a? -  ?/  a*-2  6*     a*+aM— 6b* 


Set  F.     Quadratic  Equations. 
Solve  the  following  equations: 

1.    (a; +  !)(*;+ 3)  =12+(a>  +  7)  V2. 


2.  V5  x2  -  3  a;  -  41  =  3  a;  -  7. 

3.  a/8  r3  -  35  r2  +  55  r  -  57  =  2  r  -  3. 
4 


4.   3VaT-^i  - — = —  =  4. 
Vx-1 

5     28  (3  m +  10)  25        _Q 

8  m3 -27         m(2ro-3) 

6.    2V3^  +  4  +  3V37+7=        8 

V3*  +  4 


7. 


SUPPLEMENTARY   TOPICS  281 

2  s2  -4s-3  =  s2-4s  +  2 
2s2-»2s  +  3~s2-3s  +  2' 

(See  Example  1,  Set  C.) 


g     m+1      m  +  2      ^  +  3  =  3> 
m  —  1      m  —  2      m  —  3 

Solve  the  following  equations  for  x. 

9.    or2  —  m2nx  +  mn2cc  =  m3w3. 
10.   tf-kax—  10  x  =  —  40  a. 
2a 


11.  Va  +  #-V2 

Va  +  <c 

12.  (a  +  a)3  -f  (6  -  xf  =  (a  +  &)3. 

13.  x2  —  (m  —p)  x  +  (m  —  »)  (n  —p)  =  0. 

14.  (a  +  fc)a2  +  (3a  +  &)a  =  -2a. 
15        ^         a  +  P  =  2(q2  +  &2) 

a  +  b         x  a?—b2 

**     x2-l_   4a&  -      2x  +  l       2n  +  l 

lb.      —  — — .  .  17. 


»a  -  &2  V«  +  1      Vw  +  1 

18.  a2c2  (1  +  #)2  -  &V?2  (1  -  a)2  =  0. 

19.  VS  +  s/(m  —  w)#  +  m»  =  2m. 

20.  i-  i  +i+ 4- =o. 

»     6  —  a      a     a  +  o 


INDEX 


A,  the  symbol,  35. 

Abscissa,  46. 

Absolute  value,  2. 

Antecedent,  206. 

Arithmetic,  means,  189  ;  progression, 

187. 
Ascending  powers,  6. 
Axis,  horizontal,  46 ;  vertical,  46. 

Base,  3. 

Binomial,  6  ;  square  of  a,  14  ;  theorem, 

202. 
Braces,  7. 
Brackets,  7. 

Cancellation,  in  an  equation,  37. 

Changing  signs,  in  an  equation,  37 ; 
in  a  fraction,  24. 

Characteristic,  172. 

Clearing  of  fractions,  38. 

Coefficient,  6 ;  detached,  261 ;  numer- 
ical, 7. 

Combinations,  235. 

Common,  difference,  187 ;  logarithm, 
171. 

Complex  number,  98. 

Conditional  equation,  33. 

Consequent,  206. 

Coordinates,  46. 

Cube  root,  256. 

D,  the  symbol,  35. 

Degree,    of   a   monomial,   22 ;   of   an 

equation,  49. 
Descending  powers,  6. 
Determinant,  239 ;  of  second  order, 

239  ;  of  nth  order,  244. 
Discriminant,   138. 
Division,  synthetic,  110,  272. 

Elimination,  by  addition  or  sub- 
traction, 53  ;  by  substitution,  53. 


Ellipse,  119. 

Equation,  33  ;  cancelling  terms  in  an, 
37 ;  changing  signs  in  an,  37 ; 
complete  quadratic,  78 ;  condi- 
tional, 33 ;  degree  of,  49 ;  graph- 
ical solution  of  an,  80 ;  homoge- 
neous, 126 ;  identical,  33  ;  indeter- 
minate, 48  ;  linear,  49  ;  irrational, 
166 ;  members  of  an,  33 ;  pure 
quadratic,  74  ;  rational,  126  ;  solv- 
ing an,  33  ;  transposition  in  an,  37. 

Equations,  equivalent,  36  ;  formation 
of,  137 ;  inconsistent,  51 ;  inde- 
pendent, 51 ;  simultaneous,  51, 
124  ;  system  of,  124. 

Equivalent  systems,  127. 

Evolution,  4,  142. 

Exponent,  3  ;  fractional,  144 ;  nega- 
tive, 145 ;  zero,   144. 

Exponents,  law  of  division  of,  12 ; 
law  of  multiplication  of,  10 ;  laws 
of,  140. 

Expression,  6. 

Extremes,  207. 

Factor,  1 ;  common,  6 ;  highest  com- 
mon, 22,  262  ;  to,  17  ;  theorem,  112. 

Factors,  prime,  17. 

Fourth  proportional,  208. 

Fractions,  24 ;  clearing  of,  38 ; 
equivalent,  25. 

Fundamental  operations,  3. 

Geometric,  mean,  197 ;  progression, 
195. 

Graph  of  an  equation  with  two  varia- 
bles, 117. 

Graphical  representation,  46. 

Graphical  solution  of  an  equation 
with  one  variable,  80. 

Grouping,  symbols  of,  7 ;  factoring 
by,  105. 


283 


284 


INDEX 


Homogeneous  equations,  126. 
Horizontal  axis,  46. 
Hyperbola,  120. 

Identity,  33. 

Imaginary  number,  96. 

Imaginary  numbers,  addition  and 
subtraction  of,  97 ;  multiplication 
of,  163 ;  division  of,  164. 

Imaginary  roots  in  a  quadratic  equa- 
tion, 96  ;  meaning  of,  on  graphs,  99. 

Imaginary  unit,  96. 

Inconsistent  equations,  51. 

Independent  equations,  51. 

Indeterminate  equations,  48 ;  forms, 
266. 

Index,  142. 

Infinite  geometric  progression,  199. 

Infinity,  267. 

Inversion,  243. 

Involution,  141. 

Irrational,  equation,  166 ;  number, 
138. 

Left  member  of  an  equation,  33. 
Like  terms,  7. 
Linear  equation,  49. 
Logarithm,  170;   common,  171. 
Lowest  common  multiple,  22. 

M,  the  symbol,  35. 
Mantissa,  172. 
Means,  207. 

Members  of  an  equation,  33. 
Monomial,  6. 

Monomials,  addition  of,  7 ;  division 
of,  12  ;  multiplication  of,  10. 

Negative  exponent,  145. 

Negative   numbers,    2 ;    addition   of, 

2 ;    division   of,    3 ;    multiplication 

of,  3  ;  subtraction  of,  3. 
Number,    complex,    98 ;    imaginary, 

96,  138;  irrational,  138;  negative, 

2  ;  positive,  2  ;  prime,  17 ;  rational, 

2  ;  real,  96,  138. 
Numerical  coefficient,  7. 

Order,  of  determinant,  244;  of  rad- 
ical, 150. 


Ordinate,  46. 
Origin,  46. 


Parabola,  117. 

Parentheses,  7 ;  inclosing  terms  in, 
8 ;  removing,  8. 

Perfect  square'trinomial,  17. 

Periods,  67. 

Permutations,  231. 

Polynomial,  6 ;  arranging  a,  6. 

Polynomials,  addition  of,  7  ;  division 
of,  12 ;  multiplication  of,  10 ; 
square  root  of,  64 ;  subtraction 
of,  7 ;  factoring  of,  103. 

Power,  3. 

Powers,  ascending,  6 ;  descending,  6. 

Prime  number,  17. 

Progression,  arithmetic,  187;  geo- 
metric, 195. 

Proportion,  207;  by  alternation,  210; 
by  composition,  210 ;  by  division, 
210 ;  by  composition  and  division, 
210;  by  inversion,  211. 

Proportional,  fourth,  208 ;  mean, 
208 ;  third,  208. 

Pure  quadratic,  74. 


Quadratic  equation,  74 ;  solution  of, 
by  completing  the  square,  83  ; 
by  factoring,  78 ;  by  formula,  87 ; 
complete,  78 ;  graph  of,  80 ;  imag- 
inary roots  of  a,  96 ;  pure,  74  ; 
theory  of,  135. 

Quadratic  surd,  71. 

Radical,  150 ;  index  of,  4 ;  order  of, 
150. 

Radicals,  similar,  154. 

Radicand,  4,  142. 

Ratio,  206 ;  of  a  geometric  progres- 
sion,  195. 

Rational  and  integral,  17. 

Rational  number,  2. 

Rationalizing  the  denominator,   161. 

Remainder  theorem,  109. 

Root,  cube,  256 ;  square,  of  a  frac- 
tion, 71 ;  of  an  equation,  33 ; 
principal,  142. 


INDEX 


285 


Roots,    imaginary,    of    a    quadratic, 
96,  138. 

S,  the  symbol,  35. 

Signs,    change    of,    in    an    equation, 

37 ;    law    of,    in    addition,    2 ;    in 

division,      3 ;      in     multiplication, 

3 ;  in  subtraction,  3 ;  in  a  fraction, 

24. 
Similar  terms,  7. 
Simultaneous  equations,  51,  124. 
Square  roots,    approximate,    69 ;   by 

division,    64 ;    by    inspection,    63 ; 

of  a  number,  66 ;  of  a  polynomial, 

64 ;  of  a  trinomial,  17. 
Surd,  conjugate,  160 ;  quadratic,  71 ; 

addition  of,  72,  154. 
Symbols  of  grouping,  7. 


Synthetic  division,  110,  272. 
System  of  equations,  124. 

Table  of  square  roots,  70;  of  loga- 
rithms, 174. 

Term,  6  ;  degree  of,  49. 

Terms,  dissimilar,  7  ;  like,  7  ;  similar, 
7 ;  unlike,  7. 

Theorem,  binomial,  202  ;  factor,  112  ; 
remainder,  109. 

Transposition,  37. 

Unit,  imaginary,  96. 

Unlike  terms,  7. 

Variables,  48. 
Vertical  axis,  46. 
Vinculum,  7. 


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